Title: Derivatives
Course: Demonstrating STACK (v4.4.7)

Find the derivative of the following function: \[ y=[value] \sin^{-1}\left( [value] [value]^2\right) \]
Feedback: In order to solve this problem we will have to use the chain rule \(\frac{df}{dx}=\frac{df}{dg} \cdot\frac{dg}{dx}\) as follows: Define \(g([value])=[value][value]\) and \(f([value])=[value] \sin^{-1}\left( g\right)\). So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}\left([value] \sin^{-1}\left( [value] [value]^2\right)\right) &\displaystyle =&\displaystyle \frac{df}{dg}\cdot \frac{dg}{d[value]} \\ &\displaystyle =&\displaystyle \frac{d}{dg}([value] \sin^{-1}\left( g\right))\cdot \frac{d}{d[value]}([value] [value]^2) \\ &\displaystyle =&\displaystyle \frac[value]{ \sqrt{1-g^2}}\cdot [value][value] \\ &\displaystyle =&\displaystyle \frac{[value][value]}{ \sqrt{1-[value][value]}} \end{array}\]

Find \[\frac{dy}{dx}\] when \[y=\cosh^{-1}\left(\sqrt{[value] [value] + [value]}\right) .\]
Feedback: To find \(\frac{dy}{dx}\) when \[y=\cosh^{-1}\left(\sqrt{[value] [value] + [value]}\right) .\] In order to solve this problem we will have to use the chain rule \(\frac{df}{d[value]}=\frac{df}{dg} \cdot\frac{dg}{dh}\cdot\frac{dh}{d[value]}\) as follows: Define \(h([value])=[value][value]+[value]\), \(g(h) = \sqrt{h} = h^{\frac{1}{2}}\) and \(f(g)=\cosh^{-1}\left(g\right)\). So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}\left(\cosh^{-1}\left(\sqrt{[value] [value] + [value]}\right)\right) &\displaystyle =&\displaystyle \frac{df}{dg}\cdot \frac{dg}{dh}\cdot \frac{dh}{d[value]} \\ &\displaystyle =&\displaystyle \frac{d}{dg}(\cosh^{-1}\left(g\right))\cdot\frac{d}{dh}(h^{\frac{1}{2}})\cdot \frac{d}{d[value]}([value] [value]+[value]) \\ &\displaystyle =&\displaystyle \frac{1}{ \sqrt{g^2-1}}\cdot\frac{1}{2}(h^{-\frac{1}{2}})\cdot [value] \\ &\displaystyle =&\displaystyle \frac{1}{ \sqrt{h-1}}\cdot\frac[value]{2 \sqrt{h}} \\&\displaystyle =&\displaystyle [value]\end{array}\]

Find the derivative of the following function evaluated at the point \(c\): \[ f([value])=[value] \frac[value]{\sin([value])} \] where \[c=[value]\]
Feedback: In order to solve this problem we will have to use the product rule \(\frac{d}{dx}(fg) = \frac{df}{dx}\cdot g+f\cdot\frac{dg}{dx}\) as follows: Let \(f(x) = [value][value]\mbox{ and }g(x) = \displaystyle\frac{1}{\sin([value])}\) \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}\left([value]\right) &=& \left(\frac{d}{d[value]}([value])\right)\cdot [value] + ([value])\cdot \left(\frac{d}{d[value]}\left([value]\right)\right) \\ & =& ([value])\cdot [value] + ([value])\cdot [value] \\ &=& [value] \end{array}\] Evaluated at [value] = [value] this yields [value].

Find \[ \frac{dy}{dx}\] where \[ y(x)= [value]\]
Feedback: In order to solve this problem we will have to use the product rule \(\frac{d}{dx}(fg) = \frac{df}{dx}\cdot g+f\cdot\frac{dg}{dx}\) as follows: Let \(f(x) = \sin([value])\mbox{ and }g(x) = \displaystyle\frac{1}{\cos([value])+[value]}\) \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}\left([value]\right) & =& \frac{d}{d[value]}([value])\cdot\left([value]\right)+\frac{d}{d[value]}\left([value]\right)\cdot([value]) \\ & =& ([value])\cdot[value]+([value])\cdot[value] \\ & =& = [value]+[value] \end{array}\]

Find \[ \frac{dy}{dx}\] when \[ y = \sqrt{[value][value]}\ln([value][value])\]
Feedback: In order to solve this problem we will have to use the product rule \(\frac{d}{dx}(fg) = \frac{df}{dx}\cdot g+f\cdot\frac{dg}{dx}\) as follows: Let \(f(x) =\displaystyle \sqrt{[value][value]}\mbox{ and }g(x) = \ln([value][value])\) \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}\left(\sqrt{[value][value]}\ln([value][value])\right) & =\displaystyle & \left(\frac{d}{d[value]}(\sqrt{[value][value]})\right)\cdot \ln([value][value]) + (\sqrt{[value][value]})\cdot \left(\frac{d}{d[value]}\left(\ln([value][value])\right)\right) \\ & =& \left(\frac{\sqrt[value]}{2\sqrt [value]}\right)\cdot \ln([value][value]) + (\sqrt{[value][value]})\cdot \frac{1}[value] \\ & =& [value] \end{array}\]

Use implicit differentiation to find  \[\frac{dy}{dx}\]
for the graph of  \[x^[value]+y^[value]=[value]xy\]
Feedback: To find \[\frac{dy}{dx}\] for the graph of \[x^[value]+y^[value]=[value]xy\] rearrange the equation to give \[x^[value]+y^[value]-[value]xy =0\] Thus \[\frac{d}{dx}\left(x^[value]+y^[value]-[value]xy\right) =0\] \[[value][value]+[value][value]\frac{dy}{dx}-[value](y+x\frac{dy}{dx})=0\] \[\frac{dy}{dx}([value][value]-[value]x) = [value]y-[value][value]\] \[\frac{dy}{dx} = \frac{[value]y-[value][value]}{[value][value]-[value]x}\]