Title: Default for 18.2 Partial Derivatives
Course: HELM Workbooks 2023-2024

Partial Derivatives

When a function of more than one independent input variable changes because of changes in one or more of the input variables, it is important to calculate the change in the function itself. This can be investigated by holding all but one of the variables constant and finding the rate of change of the function with respect to the one remaining variable. This process is called partial differentiation. In this Section we show how to carry out the process.

First partial derivatives

The \(x\) partial derivative

For a function of a single variable, \(y=f(x)\), changing the independent variable \(x\) leads to a corresponding change in the dependent variable \(y\). The rate of change of \(y\) with respect to \(x\) is given by the derivative, written \(\dfrac{\mathrm{d}f}{\mathrm{d}x}\). A similar situation occurs with functions of more than one variable. For clarity we shall concentrate on functions of just two variables. In the relation \(z=f(x,y)\), the independent variables are \(x\) and \(y\), and the dependent variable is \(z\). We have seen in Section 18.1 that as \(x\) and \(y\) vary the \(z\)-value traces out a surface. Now both of the variables \(x\) and \(y\) may change simultaneously inducing a change in \(z\). However, rather than consider this general situation, to begin with we shall hold one of the independent variables fixed. This is equivalent to moving along a curve obtained by intersecting the surface by one of the coordinate planes.

Consider \(f(x,y)=x^3+2x^2y+y^2+2x+1.\)

Suppose we keep \(y\) constant and vary \(x\); then what is the rate of change of the function \(f\)?

Suppose we hold \(y\) at the value \(3\) then

\[\begin{aligned} f(x,3) &= x^3+6x^2+9+2x+1 \\ &= x^3+6x^2+2x+10 \end{aligned}\]

In effect, we now have a function of \(x\) only. If we differentiate it with respect to \(x\) we obtain the expression: \(3x^2+12x+2\).

We say that \(f\) has been partially differentiated with respect to \(x\). We denote the partial derivative of \(f\) with respect to \(x\) by \(\dfrac{\partial f}{\partial x}\) (to be read as ‘partial dee \(f\) by dee \(x\)’ ). In this example, when \(y=3\):

\[\dfrac{\partial f}{\partial x} = 3x^2+12x+2\]

In the same way if \(y\) is held at the value 4 then \(f(x,4)=x^3+8x^2+16+2x+1=x^3+8x^2+2x+17\) and so, for this value of \(y\),

\[\dfrac{\partial f}{\partial x} = 3x^2+16x+2.\]

Now if we return to the original formulation \(f(x,y)=x^3+2x^2y+y^2+2x+1\) and treat \(y\) as a constant then the process of partial differentiation with respect to \(x\) gives

\[ \begin{aligned} \frac{\partial f}{\partial x} &= 3x^2+4xy+0+2+0 \\ &= 3x^2+4xy+2. \end{aligned}\]

Key Point

The Partial Derivative of \(f\) with respect to \(x\)

For a function of two variables \(z=f(x,y)\) the partial derivative of \(f\) with respect to \(x\) is denoted by \(\dfrac{\partial f}{\partial x}\) and is obtained by differentiating \(f(x,y)\) with respect to \(x\) in the usual way but treating the \(y\)-variable as if it were a constant.

Alternative notations for \(\dfrac{\partial f}{\partial x}\) are \(f_x(x,y)\) or \(f_x\) or \(\dfrac{\partial z}{\partial x}\)

Example

Find \(\dfrac{\partial f}{\partial x}\) for

1. \(f(x,y)=x^3+x+y^2+y\),
2. \(f(x,y)=x^2y+xy^3\).

Solution

1. \(\dfrac{\partial f}{\partial x}=3x^2+1+0+0=3x^2+1\)
2. \(\dfrac{\partial f}{\partial x}=2x\cdot y+1\cdot y^3=2xy+y^3\)

Q: Exercise

For the following functions find \(\dfrac{\partial f}{\partial x}\) and \(\dfrac{\partial f}{\partial y}\)

1. \(f(x,y)=[value]\)

\(\dfrac{\partial f}{\partial x} =\)
\(\dfrac{\partial f}{\partial y} =\)
2. \(f(x,y)=[value]\)

\(\dfrac{\partial f}{\partial x} =\)
\(\dfrac{\partial f}{\partial y} =\)
3. \(f(x,y)=[value]\)

\(\dfrac{\partial f}{\partial x} =\)
\(\dfrac{\partial f}{\partial y} =\)
4. \(f(x,y)=[value]\)

\(\dfrac{\partial f}{\partial x} =\)
\(\dfrac{\partial f}{\partial y} =\)
5. \(f(x,y,z)=[value]\)

\(\dfrac{\partial f}{\partial x} =\)
\(\dfrac{\partial f}{\partial y} =\)
Feedback: 1. \(f(x,y)=[value]\) \(\dfrac{\partial f}{\partial x} = [value]\) \(\dfrac{\partial f}{\partial y} = [value]\) 2. \(f(x,y)=[value]\) \(\dfrac{\partial f}{\partial x} = [value]\) \(\dfrac{\partial f}{\partial y} = [value]\) 3. \(f(x,y)=[value]\) \(\dfrac{\partial f}{\partial x} = [value]\) \(\dfrac{\partial f}{\partial y} = [value]\) 4. \(f(x,y)=[value]\) \(\dfrac{\partial f}{\partial x} = [value]\) \(\dfrac{\partial f}{\partial y} = [value]\) 5. \(f(x,y)=[value]\) \(\dfrac{\partial f}{\partial x} = [value]\) \(\dfrac{\partial f}{\partial y} = [value]\)
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Q: Exercise

Find the value of the following partial derivatives at the points given \(f_x(1,1),\ f_x(-1,-1),\ f_y(1,2),\ f_y(2,1).\)

\[f(x,y)=[value]\]

1. \(f_{x}([value], [value]) = \)
2. \(f_{x}([value], [value]) = \)
3. \(f_{y}([value], [value]) = \)
4. \(f_{y}([value], [value]) = \)
Feedback: We are given the function \(f(x,y)=[value]\) and we are asked to evaluate \(f_x(1,1),\ f_x(-1,-1),\ f_y(1,2),\ f_y(2,1).\) First we calculate the partial derivatives: \(f_x= [value] \quad \text{and} \quad f_y= [value].\) So: 1. \(f_{x}([value], [value]) = [value]\) 2. \(f_{x}([value], [value]) = [value]\) 3. \(f_{y}([value], [value]) = [value]\) 4. \(f_{y}([value], [value]) = [value]\)
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Q: Exercise

Find the first partial derivatives of the function

\[f(x,y) = [value].\]

\(\displaystyle \frac{\partial f}{\partial x} = \)

\(\displaystyle \frac{\partial f}{\partial y} = \)

Now find \(f_x([value], [value])\), \(f_x([value], [value])\), \(f_y([value], [value])\) and \(f_y([value], [value])\).

\(f_x([value], [value]) = \)

\(f_x([value], [value]) = \)

\(f_y([value], [value]) = \)

\(f_y([value], [value]) = \)
Feedback: We are asked to find the first partial derivatives of the function \[f(x,y) = [value].\] This gives: \[ \frac{\partial f}{\partial x} = [value] \quad \text{and} \quad \frac{\partial f}{\partial y} = [value] \] Next, we are asked to evaluate these derivatives at different points. \[\begin{aligned} f_x([value], [value]) &= [value] \\ f_x([value], [value]) &= [value] \\ f_y([value], [value]) &= [value] \\ f_y([value], [value]) &= [value] \end{aligned}\]
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Q: Exercise

For the function \(f(x,y)=[value]\), evaluate the following partial derivatives at the points given \(f_x(1,1),\ f_x(-1,-1),\ f_y(1,2),\ f_y(2,1)\)

1. \(f_{x}([value], [value]) = \)
2. \(f_{x}([value], [value]) = \)
3. \(f_{y}([value], [value]) = \)
4. \(f_{y}([value], [value]) = \)
Feedback: We are given the function \(f(x,y)=[value]\) and we are asked to evaluate \(f_x(1,1),\ f_x(-1,-1),\ f_y(1,2),\ f_y(2,1).\) First we calculate the partial derivatives: \(f_x= [value] \quad \text{and} \quad f_y= [value].\) So: 1. \(f_{x}([value], [value]) = [value]\) 2. \(f_{x}([value], [value]) = [value]\) 3. \(f_{y}([value], [value]) = [value]\) 4. \(f_{y}([value], [value]) = [value]\)
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Q: Exercise

Evaluate \(f_x(1,1),\ f_x(-1,-1),\ f_y(1,2),\ f_y(2,1)\) for the function

\(f(x,y)=[value]\)

1. \(f_{x}([value], [value]) = \)
2. \(f_{x}([value], [value]) = \)
3. \(f_{y}([value], [value]) = \)
4. \(f_{y}([value], [value]) = \)
Feedback: We are given the function \(f(x,y)=[value]\) and we are asked to evaluate \(f_x(1,1),\ f_x(-1,-1),\ f_y(1,2),\ f_y(2,1).\) First we calculate the partial derivatives: \(f_x= [value] \quad \text{and} \quad f_y= [value].\) So: 1. \(f_{x}([value], [value]) = [value]\) 2. \(f_{x}([value], [value]) = [value]\) 3. \(f_{y}([value], [value]) = [value]\) 4. \(f_{y}([value], [value]) = [value]\)
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Second partial derivatives

Performing two successive partial differentiations of \(f(x,y)\) with respect to \(x\) (holding \(y\) constant) is denoted by \(\dfrac{\partial^2 f}{\partial x^2}\) (or \(f_{xx}(x,y)\)) and is defined by \(\dfrac{\partial^2 f}{\partial x^2}\equiv \dfrac{\partial}{\partial x}\left(\dfrac{\partial f}{\partial x}\right)\)

For functions of two or more variables as well as \(\dfrac{\partial^2 f}{\partial x^2}\) other second-order partial derivatives can be obtained. Most obvious is the second derivative of \(f(x,y)\) with respect to \(y\) is denoted by \(\dfrac{\partial^2 f}{\partial y^2}\) (or \(f_{yy}(x,y)\)) which is defined as: \(\dfrac{\partial^2 f}{\partial y^2}\equiv \dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial y}\right)\)

Example

Find \(\dfrac{\partial^2f}{\partial x^2}\) and \(\dfrac{\partial^2f}{\partial y^2}\) for \(f(x,y)=x^3+x^2y^2+2y^3+2x+y\).

Solution

\(\dfrac{\partial f}{\partial x}=3x^2+2xy^2+0+2+0=3x^2+2xy^2+2\)
\(\dfrac{\partial^2f}{\partial x^2}\equiv \dfrac{\partial}{\partial x}\left(\dfrac{\partial f}{\partial x}\right)=6x+2y^2+0=6x+2y^2.\)
\(\dfrac{\partial f}{\partial y}=0+x^2\times 2y+6y^2+0+1=2x^2y+6y^2+1\)
\(\dfrac{\partial^2f}{\partial y^2}=\dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial y}\right)=2x^2+12y.\)

We can use the alternative notation when evaluating derivatives.

Example

Find \(f_{xx}(-1,1)\) and \(f_{yy}(2,-2)\) for \(f(x,y)=x^3+x^2y^2+2y^3+2x+y\).

Solution

\(f_{xx}(-1,1)=6\times (-1)+2\times (-1)^2=-4.\)
\(f_{yy}(2,-2)=2\times (2)^2+12\times (-2)=-16\)

Mixed second derivatives

It is possible to carry out a partial differentiation of \(f(x,y)\) with respect to \(x\) followed by a partial differentiation with respect to \(y\) (or vice-versa). The results are examples of mixed derivatives. We must be careful with the notation here. We use \(\dfrac{\partial^2f}{\partial x\partial y}\) to mean ‘differentiate first with respect to \(y\) and then with respect to \(x\)’ and we use \(\dfrac{\partial^2f}{\partial y\partial x}\) to mean ‘differentiate first with respect to \(x\) and then with respect to \(y\)’:

i.e.\(\dfrac{\partial^2f}{\partial x\partial y}\equiv\dfrac{\partial}{\partial x}\left(\dfrac{\partial f}{\partial y}\right)\) and \(\dfrac{\partial^2f}{\partial y\partial x}\equiv\dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial x}\right).\)

(This explains why the order is opposite of what we expect - the derivative ‘operates on the left’.)

Example

For \(f(x,y)=x^3+2x^2y^2+y^3\) find \(\dfrac{\partial^2f}{\partial x\partial y}.\)

Solution

\(\dfrac{\partial f}{\partial y}=4x^2y+3y^2;\\ \dfrac{\partial^2f}{\partial x\partial y}=8xy\)

The remaining possibility is to differentiate first with respect to \(x\) and then with respect to \(y\) i.e. \(\dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial x}\right)\).

For the function in the Example above \(\dfrac{\partial f}{\partial x}=3x^2+4xy^2\) and \(\dfrac{\partial^2 f}{\partial y\partial x}=8xy.\) Notice that for this function \(\dfrac{\partial^2f}{\partial x\partial y}\equiv \dfrac{\partial^2f}{\partial y\partial x}.\) This equality of mixed derivatives is true for all functions which you are likely to meet in your studies. To evaluate a mixed derivative we can use the alternative notation. To evaluate \(\dfrac{\partial^2f}{\partial x\partial y}\) we write \(f_{yx}(x,y)\) to indicate that the first differentiation is with respect to \(y\). Similarly, \(\dfrac{\partial^2f}{\partial y\partial x}\) is denoted by \(f_{xy}(x,y)\).

Example

Find \(f_{yx}(1,2)\) for the function \(f(x,y)=x^3+2x^2y^2+y^3\)

Solution

\(f_x = 3x^2+4xy^2\)

and \(f_{yx} = 8xy\)

so \(f_{yx}(1,2)=8\times 1\times 2=16.\)

Example

Find \(f_{xx}(1,2),\ f_{yy}(-2,-1),\) \(f_{xy}(3,3)\) for \(f(x,y)\equiv x^3+3x^2y^2+y^2.\)

Solution

\(f_x=3x^2+6xy^2;\ \ f_y=6x^2y+2y\)

\(f_{xx}=6x+6y^2;\ \ f_{yy}=6x^2+2;\ \ f_{xy}=f_{yx}=12xy\)

\(f_{xx}(1,2)=6+24=30;\ \ f_{yy}(-2,-1)=26;\ \ f_{xy}(3,3)=108\)

Engineering example

The ideal gas law and Redlich-Kwong equation

Introduction

In Chemical Engineering it is often necessary to be able to equate the pressure, volume and temperature of a gas. One relevant equation is the ideal gas law \(PV = n R T\)   (1) where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles of gas, \(T\) is temperature and \(R\) is the ideal gas constant (= 8.314 J mol\(^{-1}\) K\(^{-1}\), when all quantities are in S.I. units).

The ideal gas law has been in use since 1834, although its special cases at constant temperature (Boyle’s Law, 1662) and constant pressure (Charles’ Law, 1787) had been in use many decades previously. While the ideal gas law is adequate in many circumstances, it has been superseded by many other laws where, in general, simplicity is weighed against accuracy.

One such law is the Redlich-Kwong equation \(P = \dfrac{R T}{V-b} - \dfrac{a}{\sqrt{T} V (V+b)}\)   (2) where, in addition to the variables in the ideal gas law, the extra parameters \(a\) and \(b\) are dependent upon the particular gas under consideration. Clearly, in both equations the temperature, pressure and volume will be positive. Additionally, the Redlich-Kwong equation is only valid for values of volume greater than the parameter \(b\) - in practice however, this is not a limitation, since the gas would condense to a liquid before this point was reached.

Problem in words Show that for both Equations (1) and (2)

1. for constant temperature, the pressure decreases as the volume increases (Note : in the Redlich-Kwong equation, assume that \(T\) is large.)
2. for constant volume, the pressure increases as the temperature increases.

Mathematical statement of problem

For both Equations (1) and (2), and for the allowed ranges of the variables, show that

1. \(\dfrac{\partial P}{\partial V} < 0\) for \(T\) = constant
2. \(\dfrac{\partial P}{\partial T} > 0\) for \(V\) = constant

Assume that \(T\) is sufficiently large so that terms in \(T^{-1/2}\) may be neglected when compared to terms in \(T\).

Mathematical analysis

1. Ideal gas law
This can be rearranged as \(P = \dfrac{n R T}{V}\) so that
  1. at constant temperature \(\dfrac{\partial P}{\partial V} = \dfrac{-n R \, T}{V^2} < 0\) as all quantities are positive
  2. for constant volume \(\dfrac{\partial P}{\partial T} = \dfrac{n R}{V} > 0\) as all quantities are positive
2. Redlich-Kwong equation
\(P = \dfrac{RT}{V - b} - \dfrac{a}{\sqrt{T}{V(V+b)}} \\ \phantom{P}= RT(V-b)^{-1} - a T^{-1/2} (V^2+V b)^{-1}\)
so that
  1. at constant temperature \(\dfrac{\partial P}{\partial V} = -R T(V-b)^{-2} + a T^{-1/2} (V^2+V b)^{-2}(2V+b)\) which, for large \(T\), can be approximated by \( \dfrac{\partial P}{\partial V} \approx \dfrac{-R \, T}{(V-b)^2} < 0\) as all quantities are positive
  2. for constant volume \(\dfrac{\partial P}{\partial T} = R(V-b)^{-1} + \frac{1}{2} \, a \, T^{-3/2}(V^2+Vb)^{-1} > 0\) as all quantities are positive

In practice, the restriction on \(T\) is not severe, and regions in which \({\dfrac{\partial P}{\partial V} < 0}\) does not apply are those in which the gas is close to liquefying and, therefore, the entire Redlich-Kwong equation no longer applies.

Q: Exercise

Find the partial derivatives \(\dfrac{\partial^2f}{\partial x^2},\ \ \dfrac{\partial^2f}{\partial y^2},\ \ \dfrac{\partial^2f}{\partial x\partial y},\ \ \dfrac{\partial^2f}{\partial y\partial x}\) for the function \(f(x,y)=[value]\).

1. \(\dfrac{\partial^2f}{\partial x^2} =\)
2. \(\dfrac{\partial^2f}{\partial y^2} =\)
3. \(\dfrac{\partial^2f}{\partial x\partial y} =\)
4. \(\dfrac{\partial^2f}{\partial y\partial x} =\)
Feedback: We are given the function \[ f(x,y)=[value] \] and we are aske to find the partial derivatives \(\dfrac{\partial^2f}{\partial x^2},\ \ \dfrac{\partial^2f}{\partial y^2},\ \ \dfrac{\partial^2f}{\partial x\partial y},\ \ \dfrac{\partial^2f}{\partial y\partial x}.\) First we find: \[\dfrac{\partial f}{\partial x} =[value] \quad \text{and} \quad \dfrac{\partial f}{\partial x} =[value]\] So: 1. \(\dfrac{\partial^2f}{\partial x^2} = \dfrac{\partial f}{\partial x} \left( \dfrac{\partial f}{\partial x} \right) = \dfrac{\partial f}{\partial x} \left( [value] \right) = [value]\) 2. \(\dfrac{\partial^2f}{\partial y^2} = \dfrac{\partial f}{\partial y} \left( \dfrac{\partial f}{\partial y} \right) = \dfrac{\partial f}{\partial y} \left( [value] \right) = [value]\) 3. \(\dfrac{\partial^2f}{\partial x\partial y} = \dfrac{\partial f}{\partial x} \left( \dfrac{\partial f}{\partial y} \right) = \dfrac{\partial f}{\partial x} \left( [value] \right) = [value]\) 4. \(\dfrac{\partial^2f}{\partial y\partial x} = \dfrac{\partial f}{\partial y} \left( \dfrac{\partial f}{\partial x} \right) = \dfrac{\partial f}{\partial y} \left( [value] \right) = [value]\)
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Q: Exercise

For the function \(f(x,y)=[value]\) find \(\dfrac{\partial^2f}{\partial x^2},\ \ \dfrac{\partial^2f}{\partial y^2},\ \ \dfrac{\partial^2f}{\partial x\partial y},\ \ \dfrac{\partial^2f}{\partial y\partial x}.\)

1. \(\dfrac{\partial^2f}{\partial x^2} =\)
2. \(\dfrac{\partial^2f}{\partial y^2} =\)
3. \(\dfrac{\partial^2f}{\partial x\partial y} =\)
4. \(\dfrac{\partial^2f}{\partial y\partial x} =\)
Feedback: We are given the function \[ f(x,y)=[value] \] and we are aske to find the partial derivatives \(\dfrac{\partial^2f}{\partial x^2},\ \ \dfrac{\partial^2f}{\partial y^2},\ \ \dfrac{\partial^2f}{\partial x\partial y},\ \ \dfrac{\partial^2f}{\partial y\partial x}.\) First we find: \[\dfrac{\partial f}{\partial x} =[value] \quad \text{and} \quad \dfrac{\partial f}{\partial x} =[value]\] So: 1. \(\dfrac{\partial^2f}{\partial x^2} = \dfrac{\partial f}{\partial x} \left( \dfrac{\partial f}{\partial x} \right) = \dfrac{\partial f}{\partial x} \left( [value] \right) = [value]\) 2. \(\dfrac{\partial^2f}{\partial y^2} = \dfrac{\partial f}{\partial y} \left( \dfrac{\partial f}{\partial y} \right) = \dfrac{\partial f}{\partial y} \left( [value] \right) = [value]\) 3. \(\dfrac{\partial^2f}{\partial x\partial y} = \dfrac{\partial f}{\partial x} \left( \dfrac{\partial f}{\partial y} \right) = \dfrac{\partial f}{\partial x} \left( [value] \right) = [value]\) 4. \(\dfrac{\partial^2f}{\partial y\partial x} = \dfrac{\partial f}{\partial y} \left( \dfrac{\partial f}{\partial x} \right) = \dfrac{\partial f}{\partial y} \left( [value] \right) = [value]\)
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Q: Exercise

Find the partial derivatives \(\dfrac{\partial^2f}{\partial x^2},\ \ \dfrac{\partial^2f}{\partial y^2},\ \ \dfrac{\partial^2f}{\partial x\partial y},\ \ \dfrac{\partial^2f}{\partial y\partial x}\) for \(f(x,y)=[value]\).

1. \(\dfrac{\partial^2f}{\partial x^2} =\)
2. \(\dfrac{\partial^2f}{\partial y^2} =\)
3. \(\dfrac{\partial^2f}{\partial x\partial y} =\)
4. \(\dfrac{\partial^2f}{\partial y\partial x} =\)
Feedback: We are given the function \[ f(x,y)=[value] \] and we are aske to find the partial derivatives \(\dfrac{\partial^2f}{\partial x^2},\ \ \dfrac{\partial^2f}{\partial y^2},\ \ \dfrac{\partial^2f}{\partial x\partial y},\ \ \dfrac{\partial^2f}{\partial y\partial x}.\) First we find: \[\dfrac{\partial f}{\partial x} =[value] \quad \text{and} \quad \dfrac{\partial f}{\partial x} =[value]\] So: 1. \(\dfrac{\partial^2f}{\partial x^2} = \dfrac{\partial f}{\partial x} \left( \dfrac{\partial f}{\partial x} \right) = \dfrac{\partial f}{\partial x} \left( [value] \right) = [value]\) 2. \(\dfrac{\partial^2f}{\partial y^2} = \dfrac{\partial f}{\partial y} \left( \dfrac{\partial f}{\partial y} \right) = \dfrac{\partial f}{\partial y} \left( [value] \right) = [value]\) 3. \(\dfrac{\partial^2f}{\partial x\partial y} = \dfrac{\partial f}{\partial x} \left( \dfrac{\partial f}{\partial y} \right) = \dfrac{\partial f}{\partial x} \left( [value] \right) = [value]\) 4. \(\dfrac{\partial^2f}{\partial y\partial x} = \dfrac{\partial f}{\partial y} \left( \dfrac{\partial f}{\partial x} \right) = \dfrac{\partial f}{\partial y} \left( [value] \right) = [value]\)
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Q: Exercise

If \(f(x,y)=[value]\), find \(\dfrac{\partial^2f}{\partial x^2},\ \ \dfrac{\partial^2f}{\partial y^2},\ \ \dfrac{\partial^2f}{\partial x\partial y},\ \ \dfrac{\partial^2f}{\partial y\partial x}.\)

1. \(\dfrac{\partial^2f}{\partial x^2} =\)
2. \(\dfrac{\partial^2f}{\partial y^2} =\)
3. \(\dfrac{\partial^2f}{\partial x\partial y} =\)
4. \(\dfrac{\partial^2f}{\partial y\partial x} =\)
Feedback: We are given the function \[ f(x,y)=[value] \] and we are aske to find the partial derivatives \(\dfrac{\partial^2f}{\partial x^2},\ \ \dfrac{\partial^2f}{\partial y^2},\ \ \dfrac{\partial^2f}{\partial x\partial y},\ \ \dfrac{\partial^2f}{\partial y\partial x}.\) First we find: \[\dfrac{\partial f}{\partial x} =[value] \quad \text{and} \quad \dfrac{\partial f}{\partial x} =[value]\] So: 1. \(\dfrac{\partial^2f}{\partial x^2} = \dfrac{\partial f}{\partial x} \left( \dfrac{\partial f}{\partial x} \right) = \dfrac{\partial f}{\partial x} \left( [value] \right) = [value]\) 2. \(\dfrac{\partial^2f}{\partial y^2} = \dfrac{\partial f}{\partial y} \left( \dfrac{\partial f}{\partial y} \right) = \dfrac{\partial f}{\partial y} \left( [value] \right) = [value]\) 3. \(\dfrac{\partial^2f}{\partial x\partial y} = \dfrac{\partial f}{\partial x} \left( \dfrac{\partial f}{\partial y} \right) = \dfrac{\partial f}{\partial x} \left( [value] \right) = [value]\) 4. \(\dfrac{\partial^2f}{\partial y\partial x} = \dfrac{\partial f}{\partial y} \left( \dfrac{\partial f}{\partial x} \right) = \dfrac{\partial f}{\partial y} \left( [value] \right) = [value]\)
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Q: Exercise

Find the second partial derivatives of the function

\[f(x,y) = [value].\]

\(\displaystyle \frac{\partial^2 f}{\partial x^2} = \) \(\qquad \displaystyle \frac{\partial^2 f}{\partial y^2} = \)

\(\displaystyle \frac{\partial^2 f}{\partial y \partial x} = \) \(\qquad \displaystyle \frac{\partial^2 f}{\partial x \partial y} = \)

Now find \(f_{xx}([value], [value])\), \(f_{yy}([value], [value])\) and \(f_{xy}([value], [value])\).

\(f_{xx}([value], [value]) = \)

\(f_{yy}([value], [value]) = \)

\(f_{xy}([value], [value]) = \)
Feedback: We are asked to find the second partial derivatives of the function \[f(x,y) = [value].\] First, we find the first partial derivatives: \[\frac{\partial f}{\partial x} = [value] \quad \text{and} \quad \frac{\partial f}{\partial y} = [value]\] We can then find the second derivatives: \[ \frac{\partial^2 f}{\partial x^2} = [value], \qquad \frac{\partial^2 f}{\partial y^2} = [value]\] \[ \frac{\partial^2 f}{\partial y \partial x} = [value], \qquad \frac{\partial^2 f}{\partial x \partial y} = [value] \] Next, we are asked to evaluate these derivatives at different points. For this case in any point the derivatives will be \(0.\) \[\begin{aligned} f_{xx}([value], [value]) &= [value] \\ f_{yy}([value], [value]) &= [value] \\ f_{xy}([value], [value]) &= [value] \\ \end{aligned}\]
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Q: Exercise

For the following functions find \(\dfrac{\partial f}{\partial x}\) and \(\dfrac{\partial^2f}{\partial x\partial t}\)

1. \(f(x,t)=x\sin (tx)+x^2t\)

\(\dfrac{\partial f}{\partial x} = \)
\(\dfrac{\partial^2f}{\partial x\partial t} = \)
2. \(f(x,t,z)=[value]\)

\(\dfrac{\partial f}{\partial x} = \)
\(\dfrac{\partial^2f}{\partial x\partial t} = \)
3. \(f(x,t)=[value]\)

\(\dfrac{\partial f}{\partial x} = \)
\(\dfrac{\partial^2f}{\partial x\partial t} = \)
Feedback: 1. \(f(x,t)=x\sin (tx)+x^2t\) \(\dfrac{\partial f}{\partial x} = [value]\) \(\dfrac{\partial^2f}{\partial x\partial t} = [value]\) 2. \(f(x,t,z)=[value]\) \(\dfrac{\partial f}{\partial x} = [value]\) \(\dfrac{\partial^2f}{\partial x\partial t} = [value]\) 3. \(f(x,t)=[value]\) \(\dfrac{\partial f}{\partial x} = [value]\) \(\dfrac{\partial^2f}{\partial x\partial t} = [value]\)
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Q: Exercise

Find the values of the partial derivatives \(f_{xx}(1,-3),\ f_{yy}(-2,-2),\ f_{xy}(-1,3)\) if \(f(x,y)=[value]\)

1. \(f_{xx}([value], [value]) = \)
2. \(f_{yy}([value], [value]) = \)
3. \(f_{xy}([value], [value]) = \)
Feedback: We are given the function \(f(x,y)=[value]\) and we are asked to evaluate \(f_{xx}(1,-3),\ f_{yy}(-2,-2),\ f_{xy}(-1,3).\) First we calculate the partial derivatives: \[f_x= [value] \quad \text{and} \quad f_y= [value]\] \[ f_{xx}= [value], \quad f_{yy}=[value], \quad f_{xy}=[value] \] So: 1. \(f_{xx}([value], [value]) = [value]\) 2. \(f_{yy}([value], [value]) = [value]\) 3. \(f_{xy}([value], [value]) = [value]\)
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Q: Exercise

For the function \([value]\) find the values of the following partial derivatives \(f_{xx}(1,-3),\ f_{yy}(-2,-2),\ f_{xy}(-1,3)\).

1. \(f_{xx}([value], [value]) = \)
2. \(f_{yy}([value], [value]) = \)
3. \(f_{xy}([value], [value]) = \)
Feedback: We are given the function \(f(x,y)=[value]\) and we are asked to evaluate \(f_{xx}(1,-3),\ f_{yy}(-2,-2),\ f_{xy}(-1,3).\) First we calculate the partial derivatives: \[f_x= [value] \quad \text{and} \quad f_y= [value]\] \[ f_{xx}= [value], \quad f_{yy}=[value], \quad f_{xy}=[value] \] So: 1. \(f_{xx}([value], [value]) = [value]\) 2. \(f_{yy}([value], [value]) = [value]\) 3. \(f_{xy}([value], [value]) = [value]\)
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Q: Exercise

For the following function find \(f_{xx}(1,-3),\ f_{yy}(-2,-2),\ f_{xy}(-1,3)\)

\(f(x,y)=[value]\)

1. \(f_{xx}([value], [value]) = \)
2. \(f_{yy}([value], [value]) = \)
3. \(f_{xy}([value], [value]) = \)
Feedback: We are given the function \(f(x,y)=[value]\) and we are asked to evaluate \(f_{xx}(1,-3),\ f_{yy}(-2,-2),\ f_{xy}(-1,3).\) First we calculate the partial derivatives: \[f_x= [value] \quad \text{and} \quad f_y= [value]\] \[ f_{xx}= [value], \quad f_{yy}=[value], \quad f_{xy}=[value] \] So: 1. \(f_{xx}([value], [value]) = [value]\) 2. \(f_{yy}([value], [value]) = [value]\) 3. \(f_{xy}([value], [value]) = [value]\)
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Exercise

Find the first partial derivatives with respect to \(x\) and \(y\) of the function

\[f(x,y,z) = [value].\]

\(\displaystyle \frac{\partial f}{\partial x} = \)

\(\displaystyle \frac{\partial f}{\partial y} = \)
Feedback: We are asked to find the first partial derivatives with respect to \(x\) and \(y\) of the function \[f(x,y) = [value].\] This gives: \[ \frac{\partial f}{\partial x} = [value] \quad \text{and} \quad \frac{\partial f}{\partial y} = [value] \]

Q: Exercise

Find the second partial derivatives of the function

\[f(x,y) = [value].\]

\(\displaystyle \frac{\partial^2 f}{\partial x^2} = \) \(\qquad \displaystyle \frac{\partial^2 f}{\partial y^2} = \)

\(\displaystyle \frac{\partial^2 f}{\partial y \partial x} = \) \(\qquad \displaystyle \frac{\partial^2 f}{\partial x \partial y} = \)

Now find \(f_{xx}([value], [value])\), \(f_{yy}([value], [value])\) and \(f_{xy}([value], [value])\).

\(f_{xx}([value], [value]) = \)

\(f_{yy}([value], [value]) = \)

\(f_{xy}([value], [value]) = \)
Feedback: We are asked to find the second partial derivatives of the function \[f(x,y) = [value].\] First, we find the first partial derivatives: \[\frac{\partial f}{\partial x} = [value] \quad \text{and} \quad \frac{\partial f}{\partial y} = [value]\] We can then find the second derivatives: \[\frac{\partial^2 f}{\partial x^2} = [value], \qquad \frac{\partial^2 f}{\partial y^2} = [value] \] \[ \frac{\partial^2 f}{\partial y \partial x} = [value], \qquad \frac{\partial^2 f}{\partial x \partial y} = [value] \] Next, we are asked to evaluate these derivatives at different points. \[\begin{aligned} f_{xx}([value], [value]) &= [value] \\ f_{yy}([value], [value]) &= [value] \\ f_{xy}([value], [value]) &= [value] \\ \end{aligned}\]
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Q: Exercise

Find the second partial derivatives of the function

\[f(x,y) = [value].\]

\(\displaystyle \frac{\partial^2 f}{\partial x^2} = \) \( \qquad \displaystyle \frac{\partial^2 f}{\partial y^2} = \)

\(\displaystyle \frac{\partial^2 f}{\partial y \partial x} = \) \( \qquad \displaystyle \frac{\partial^2 f}{\partial x \partial y} = \)

Now find \(f_{xx}([value], [value])\), \(f_{yy}([value], [value])\) and \(f_{xy}([value], [value])\).

\(f_{xx}([value], [value]) = \)

\(f_{yy}([value], [value]) = \)

\(f_{xy}([value], [value]) = \)
Feedback: We are asked to find the second partial derivatives of the function \[f(x,y) = [value].\] First, we find the first partial derivatives: \[\frac{\partial f}{\partial x} = [value] \quad \text{and} \quad \frac{\partial f}{\partial y} = [value]\] We can then find the second derivatives: \[ \frac{\partial^2 f}{\partial x^2} = [value], \qquad \frac{\partial^2 f}{\partial y^2} = [value] \] \[ \frac{\partial^2 f}{\partial y \partial x} = [value], \qquad \frac{\partial^2 f}{\partial x \partial y} = [value]\] Next, we are asked to evaluate these derivatives at different points. \[\begin{aligned} f_{xx}([value], [value]) &= [value] \\ f_{yy}([value], [value]) &= [value] \\ f_{xy}([value], [value]) &= [value] \\ \end{aligned}\]
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Q: Exercise

Find the second partial derivatives of the function

\[f(x,y) = [value].\]

\(\displaystyle \frac{\partial^2 f}{\partial x^2} = \)

\(\displaystyle \frac{\partial^2 f}{\partial y^2} = \)

\(\displaystyle \frac{\partial^2 f}{\partial y \partial x} = \)

\(\displaystyle \frac{\partial^2 f}{\partial x \partial y} = \)

Now find \(f_{xx}([value], [value])\), \(f_{yy}([value], [value])\) and \(f_{xy}([value], [value])\).

\(f_{xx}([value], [value]) = \)

\(f_{yy}([value], [value]) = \)

\(f_{xy}([value], [value]) = \)
Feedback: We are asked to find the second partial derivatives of the function \[f(x,y) = [value].\] First, we find the first partial derivatives: \[\frac{\partial f}{\partial x} = [value] \quad \text{and} \quad \frac{\partial f}{\partial y} = [value]\] We can then find the second derivatives: \[\begin{aligned} \frac{\partial^2 f}{\partial x^2} &= [value] \\ \frac{\partial^2 f}{\partial y^2} &= [value] \\ \frac{\partial^2 f}{\partial y \partial x} &= [value] \\ \frac{\partial^2 f}{\partial x \partial y} &= [value] \end{aligned}\] Next, we are asked to evaluate these derivatives at different points. \[\begin{aligned} f_{xx}([value], [value]) &= [value] \\ f_{yy}([value], [value]) &= [value] \\ f_{xy}([value], [value]) &= [value] \\ \end{aligned}\]
  Hint: (Hint: you need to simplify sin(t^3 + 1))

Exercise

Find the second partial derivatives of the function

\[f(x,y) = [value].\]

\(\displaystyle \frac{\partial^2 f}{\partial x^2} = \) \( \qquad \displaystyle \frac{\partial^2 f}{\partial y^2} = \)

\(\displaystyle \frac{\partial^2 f}{\partial y \partial x} = \) \( \qquad \displaystyle \frac{\partial^2 f}{\partial x \partial y} = \)
Feedback: We are asked to find the second partial derivatives of the function \[f(x,y) = [value].\] First, we find the first partial derivatives: \[\frac{\partial f}{\partial x} = [value] \quad \text{and} \quad \frac{\partial f}{\partial y} = [value]\] We can then find the second derivatives: \[\begin{aligned} \frac{\partial^2 f}{\partial x^2} &= [value] \\ \frac{\partial^2 f}{\partial y^2} &= [value] \\ \frac{\partial^2 f}{\partial y \partial x} &= [value] \\ \frac{\partial^2 f}{\partial x \partial y} &= [value] \end{aligned}\]

The \(y\) partial derivative

For functions of two variables \(f(x,y)\) the \(x\) and \(y\) variables are on the same footing, so what we have done for the \(x\)-variable we can do for the \(y\)-variable. We can thus imagine keeping the \(x\)-variable fixed and determining the rate of change of \(f\) as \(y\) changes. This rate of change is denoted by \(\dfrac{\partial f}{\partial y}\).

Key Point

The Partial Derivative of \(f\) with respect to \(y\)

For a function of two variables \(z=f(x,y)\) the partial derivative of \(f\) with respect to \(y\) is denoted by \(\dfrac{\partial f}{\partial y}\) and is obtained by differentiating \(f(x,y)\) with respect to \(y\) in the usual way but treating the \(x\)-variable as if it were a constant.

Alternative notations for \(\dfrac{\partial f}{\partial y}\) are \(f_y(x,y)\) or \(f_y\) or \(\dfrac{\partial z}{\partial y}\).

Returning to \(f(x,y)=x^3+2x^2y+y^2+2x+1\) once again, we therefore obtain:

\[\begin{aligned} \frac{\partial f}{\partial y} &= 0+2x^2 \cdot 1+2y+0+0 \\ &= 2x^2+2y \end{aligned}\]

Example

Find \(\dfrac{\partial f}{\partial y}\) for
1. \(f(x,y)=x^3+x+y^2+y\),
2. \(f(x,y)=x^2y+xy^3\).

Solution

1. \(\dfrac{\partial f}{\partial y}=0+0+2y+1=2y+1\)
2. \(\dfrac{\partial f}{\partial y}=x^2\cdot 1+x\cdot 3y^2=x^2+3xy^2\)

We can calculate the partial derivative of \(f\) with respect to \(x\) and the value of \(\dfrac{\partial f}{\partial x}\) at a specific point e.g. \(x=1, y=-2\).

Example

[Remember \(f_x\) means \(\dfrac{\partial f}{\partial x}\) and \(f_y\) means \(\dfrac{\partial f}{\partial y}\).]

Find \(f_x(1,-2)\) and \(f_y(-3,2)\) for \(f(x,y)=x^2+y^3+2xy\)

Solution

\(f_x(x,y)=2x+2y\), so \(f_x(1,-2)=2-4=-2\);

\(f_y(x,y)=3y^2+2x,\) so \(f_y(-3,2)=12-6=6\).

Functions of several variables

As we have seen, a function of two variables \(f(x,y)\) has two partial derivatives, \(\dfrac{\partial f}{\partial x}\) and \(\dfrac{\partial f}{\partial y}\). In an exactly analogous way a function of three variables \(f(x,y,u)\) has three partial derivatives \(\dfrac{\partial f}{\partial x}\), \(\dfrac{\partial f}{\partial y}\) and \(\dfrac{\partial f}{\partial u}\), and so on for functions of more than three variables. Each partial derivative is obtained in the same way as stated in the following Key Point:

Key Point

The Partial Derivatives of \(f(x,y,u,v,w,\dots)\)

For a function of several variables \(z=f(x, y, u, v, w, \dots)\) the partial derivative of \(f\) with respect to \(v\) (say) is denoted by \(\dfrac{\partial f}{\partial v}\) and is obtained by differentiating \(f(x, y, u, v, w, \dots)\) with respect to \(v\) in the usual way but treating all the other variables as if they were constants.

Alternative notations for \(\dfrac{\partial f}{\partial v}\) when \(z = f(x, y, u, v, w, \dots )\) are \(f_v(x, y, u, v, w \dots )\) and \(f_v\) and \(\dfrac{\partial z}{\partial v}\).

Example

Find \(\dfrac{\partial f}{\partial x}\) and \(\dfrac{\partial f}{\partial u}\) for \(f(x, y, u, v) = x^2+xy^2+y^2u^3-7uv^4\)

Solution

\(\dfrac{\partial f}{\partial x}=2x+y^2+0+0=2x+y^2\),

\(\dfrac{\partial f}{\partial u}=0+0+y^2\cdot 3u^2-7v^4=3y^2u^2-7v^4\)

Example

The pressure, \(P\), for one mole of an ideal gas is related to its absolute temperature, \(T\), and specific volume, \(v\), by the equation \(Pv=RT\) where \(R\) is the gas constant. Obtain simple expressions for

1. the coefficient of thermal expansion, \(\alpha\), defined by: \(\alpha = \dfrac{1}{v}\left(\dfrac{\partial v}{\partial T}\right)_P\)

2. the isothermal compressibility, \(\kappa_T\), defined by: \(\kappa_T=-\dfrac{1}{v}\left(\dfrac{\partial v}{\partial P}\right)_T\)

Solution

\(v=\dfrac{RT}{P}\quad\Rightarrow\quad \left(\dfrac{\partial v}{\partial T}\right)_P=\dfrac{R}{P}\)

so \(\alpha=\dfrac{1}{v}\left(\dfrac{\partial v}{\partial T}\right)_P=\dfrac{R}{Pv}=\dfrac{1}{T}\)

\(v=\dfrac{RT}{P}\quad\Rightarrow\quad \left(\dfrac{\partial v}{\partial P}\right)_T=-\dfrac{RT}{P^2}\)

so \(\kappa_T=-\dfrac{1}{v}\left(\dfrac{\partial v}{\partial P}\right)_T=\dfrac{RT}{vP^2}=\dfrac{1}{P}\)

Q: Exercise

Find the first partial derivatives of the function

\[f(x,y) = [value].\]

\(\displaystyle \frac{\partial f}{\partial x} = \)

\(\displaystyle \frac{\partial f}{\partial y} = \)

Now find \(f_x([value], [value])\) and \(f_x([value], [value]).\)

\(f_x([value], [value]) = \)

\(f_x([value], [value]) = \)
Feedback: We are asked to find the first partial derivatives of the function \[f(x,y) = [value].\] This gives: \[ \frac{\partial f}{\partial x} = [value] \quad \text{and} \quad \frac{\partial f}{\partial y} = [value] \] Next, we are asked to evaluate these derivatives at different points. \[\begin{aligned} f_x([value], [value]) &= [value] \\ f_x([value], [value]) &= [value] \end{aligned}\]
  Hint: (Hint: you need to simplify sin(t^3 + 1))