Title: CR_Diff_09
Course: Demonstrating STACK (v4.3.7)

Miscellaneous differentiation

Find the following derivative: \[ \frac{d}{d[value]}\left[ [value]\right]. \]
Feedback: The first thing to do is to rearrange the expression before we differentiate. \[ [value]=[value].\] To solve this problem you will need to use the product rule. Let \(u=[value]\) and \(v=[value]\). So, \[\begin{array}{rcl} \displaystyle \frac{d}{d[value]}[[value]] &=& \displaystyle \frac{du}{d[value]}\cdot v + u\cdot \frac{dv}{d[value]} \\ &=& \displaystyle [value] \cdot [value] + [value]\cdot[value] \\ &=& \displaystyle [value]. \end{array} \]

Find the following derivative: \[ \frac{d}{d[value]}\left[\left([value]\right)^[value]\right]. \]
Feedback: To solve this problem you will need to use the chain rule calc_chain_rule and the quotient rule calc_quotient_rule. Let \[f([value])=[value]\] and \[g([value])=[value].\] Then \(f([value])=[g([value])]^[value]\). So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}\left[[value]\right] &\displaystyle =&\displaystyle \frac{df}{dg}\cdot \frac{dg}{dx} \\ &\displaystyle =&\displaystyle \frac{d}{dg}[g^[value]]\cdot \frac{d}{d[value]}[g] \end{array}\] In order to find \(\frac{dg}{d[value]}\) you have to define two new functions: \(h([value])=[value]\) and \(i([value])=[value]\). So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}(g) &\displaystyle =&\displaystyle \frac{d}{d[value]}\left[[value]\right] \\ &\displaystyle =&\displaystyle \frac{d}{d[value]}\left(\frac{h}{i}\right) \\ &\displaystyle =&\displaystyle \frac{\frac{dh}{d[value]}\cdot i - h\cdot \frac{di}{d[value]}}{i^2} \\ &\displaystyle =&\displaystyle \frac{\left[\frac{d}{d[value]}([value])\right]\cdot ([value]) - ([value])\cdot \left[\frac{d}{d[value]}([value])\right]}{([value])^2} \\ &\displaystyle =&\displaystyle \frac{[[value]]\cdot ([value]) - ([value])\cdot [[value]]}[value] \\ &\displaystyle =&\displaystyle \frac[value][value]. \end{array}\] Now, \[\begin{array}{rcl}\displaystyle \frac{d}{dg}[g^[value]] &\displaystyle =&\displaystyle [value]g^[value] \\ &\displaystyle =&\displaystyle [value] \end{array}\] Hence, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}\left[[value]\right] &\displaystyle =&\displaystyle \frac{d}{dg}[g^[value]]\cdot \frac{d}{d[value]}[g] \\ &\displaystyle =&\displaystyle \frac[value][value]\cdot [value] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Find the following derivative:
\[ \frac{d}{d[value]}[[value]]. \]
Feedback: To solve this question, you will have to use the chain rule calc_chain_rule more than once. Let \(f([value])=[value]\) and \(g([value])=[value]\). Then, \(f([value])=e^{(g([value]))}\) \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}[[value]] &\displaystyle =&\displaystyle \frac{df}{dg}\cdot \frac{dg}{dx} \\ &\displaystyle =&\displaystyle \frac{d}{dg}(e^{g})\cdot \frac{d}{dx}[[value]] \end{array}\] To find \[\frac{d}{dx}[[value]]\] you have to define a new function: \(h([value])=[value]\), so that \(g([value])=h^[value]\). So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}[[value]] &\displaystyle =&\displaystyle \frac{dg}{dh}\cdot \frac{dh}{d[value]} \\ &\displaystyle =&\displaystyle \frac{d}{dh}(h^[value])\cdot \frac{d}{d[value]}([value]) \\ &\displaystyle =&\displaystyle [value]h^[value]\cdot ([value]) \\ &\displaystyle =&\displaystyle [value] \end{array}\] So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}[[value]] &\displaystyle =&\displaystyle \frac{d}{dg}(e^{g})\cdot \frac{d}{dx}[[value]] \\ &\displaystyle =&\displaystyle e^g \cdot [value] \\ &\displaystyle =&\displaystyle e^[value][value] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Find the following second derivative:
\[ \frac{d^2}{d[value]^2}([value]). \]
Feedback: For this question, note that the second derivative of a function is the derivative of the first derivative of that function, that is, \[\frac{d^2f}{dx^2}=\frac{d}{dx}\left(\frac{df}{dx}\right)\] To calculate the first derivative you can use the chain rule, so that \[\frac{d}{d[value]}([value]) = [value]\] So, to calculate the second derivative of [value] with respect to [value], you have to use the product rule to differentiate its first derivative with respect to [value]: \[\begin{array}{rcl}\displaystyle \frac{d^2}{d[value]^2}([value]) &\displaystyle =&\displaystyle \frac{d}{d[value]}\left[\frac{d}{d[value]}([value])\right] \\ &\displaystyle =&\displaystyle \frac{d}{d[value]}([value]) \\ &\displaystyle =&\displaystyle \left[ \frac{d}{d[value]}([value])\right]\cdot [value] + [value]\cdot \left[\frac{d}{d[value]}([value])\right] \\ &\displaystyle =&\displaystyle [value] + [value] \end{array}\]

Find the following derivative:
\[ \frac{d}{d[value]}([value]). \]
Hint: write \(\mathrm{sec}([value])\) as \(1/[value]\)
Feedback: Since \([value]=\frac{1}[value]\), \([value]=\frac{1}[value]^[value]\) So, using the chain rule [[facts:calc_chain_rule]] and the quotient rule [[facts:calc_quotient_rule]], \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}([value]) &\displaystyle =&\displaystyle \frac{d}{d[value]}\left[\left(\frac{1}[value]\right)^[value]\right] \\ &\displaystyle =&\displaystyle [value]\left[\frac{1}[value]\right]^{([value]-1)}\cdot \frac{d}{d[value]}\left[\frac{1}[value]\right] \\ &\displaystyle =&\displaystyle \frac[value]{[value]^[value]}\cdot \frac{\frac{d}{d[value]}[1]\cdot [value]-1\cdot \frac{d}{d[value]}[[value]]}{[value]^2} \\ &\displaystyle =&\displaystyle [value]\cdot \frac{0-[value]}[value] \\ &\displaystyle =&\displaystyle [value] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Find the following derivative:
\[ \frac{d}{d[value]}([value]). \]
Feedback: The simplest way to solve this problem is using the chain rule [[facts:calc_chain_rule]] by defining \(f([value])=[value]\) and \(g([value])=[value]\) so that \(f([value])=e^{g([value])}\). So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}[[value]] &\displaystyle =&\displaystyle \frac{df}{dg}\cdot \frac{dg}{dx} \\ &\displaystyle =&\displaystyle \frac{d}{dg}(e^g)\cdot \frac{d}{dx}([value]) \\ &\displaystyle =&\displaystyle e^g\cdot ([value]) \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Find the following derivative:
\[ \frac{d}{d[value]}([value]). \]
Feedback: The simplest way to solve this problem is using the product rule. [[facts:calc_product_rule]] Define \(f=[value]\) and \(g=[value]\) so that \(f'=\log([value])[value]\) and \(g'=[value][value]^[value]\). So, \[ \begin{array}{rcl} \displaystyle \frac{d}{d[value]}([value]) &\displaystyle =&\displaystyle f\cdot g' + f'\cdot g \\ &\displaystyle =&\displaystyle [value] \cdot [value][value]^[value] + [value] \\ & =&\displaystyle [value]. \end{array} \]

Find the following derivative:
\[ \frac{d}{d[value]}\left( [value]\right). \]
Feedback: To solve this problem you will need to use the product rule calc_product_rule and the quotient rulecalc_quotient_rule. Let \(f([value])=[value]\) and \(g([value])=[value]\). Then, \[\frac{d}{d[value]}\left[[value]\right] = \frac{\frac{df}{d[value]}\cdot g - f\cdot \frac{dg}{d[value]}}{g^2}.\] In order to calculate \(\frac{dg}{d[value]}\) you need to use the product rule, by defining two new functions: \(h([value])=[value]\) and \(i([value])=[value]\) so that \(g([value])=h([value])i([value])\). So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}[[value]] &\displaystyle =&\displaystyle \frac{dh}{d[value]}\cdot i + h\cdot \frac{di}{d[value]}\\ &\displaystyle =&\displaystyle [[value]][value] + [value][[value]] \\ &\displaystyle =&\displaystyle [value] \end{array}\] Hence, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}\left[[value]\right] &\displaystyle =&\displaystyle \frac{\frac{df}{d[value]}\cdot g - f\cdot \frac{dg}{d[value]}}{g^2} \\ &\displaystyle =&\displaystyle \frac{[[value]][value] - [value][[value]]}{[[value]]^2} \\ &\displaystyle =&\displaystyle \frac{[value] - [value]}[value] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Find the following derivative: \[ \frac{d}{d[value]}\left[ [value]\right]. \]
Feedback: To solve this problem you will need to use the product rule calc_product_rule and the chain rulecalc_chain_rule. Let \(g([value])=[value]\) and \(f([value])=e^{g([value])}\). Then, \[\frac{d}{d[value]}\left[[value]\right] = \frac{df}{dg}\cdot \frac{dg}{d[value]}.\] In order to calculate \(\frac{dg}{d[value]}\) you need to use the chain rule, by defining two new functions: \(h([value])=[value]\) and \(i([value])=[value]\) so that \(g([value])=h([value])i([value])\). So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}[[value]] &\displaystyle =&\displaystyle \frac{dh}{d[value]}\cdot i + h\cdot \frac{di}{d[value]}\\ &\displaystyle =&\displaystyle [[value]][value] + [value][[value]] \\ &\displaystyle =&\displaystyle [value] \end{array}\] Hence, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}\left[[value]\right] &\displaystyle =&\displaystyle \frac{df}{dg}\cdot \frac{dg}{d[value]} \\ &\displaystyle =&\displaystyle \frac{d}{dg}(e^g) \cdot \frac{d}{d[value]}([value]) \\ &\displaystyle =&\displaystyle e^[value] \cdot ([value]) \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Find the following derivative: \[ \frac{d}{d[value]}\left[ [value]\right]. \]
Feedback: To solve this problem you will need to use the product rule calc_product_rule and the chain rulecalc_chain_rule. Let \(g([value])=[value]\) and \(f([value])=g([value])^[value]\). Then, by the chain rule: \[\frac{d}{d[value]}\left[[value]\right] = \frac{df}{dg}\cdot\frac{dg}{d[value]}.\] In order to calculate \(\frac{dg}{d[value]}\) you need to use the product rule, by defining two new functions: \(h([value])=[value]\) and \(i([value])=[value]\) so that \(g([value])=h([value])i([value])\). So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}[[value]] &\displaystyle =&\displaystyle \frac{dh}{d[value]}\cdot i + h\cdot \frac{di}{d[value]}\\ &\displaystyle =&\displaystyle [[value]][value] + [value][[value]] \\ &\displaystyle =&\displaystyle [value] \end{array}\] Hence, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}\left[[value]\right] &\displaystyle =&\displaystyle \frac{df}{dg}\cdot\frac{dg}{d[value]} \\ &\displaystyle =&\displaystyle [value] \cdot [[value]] \\ &\displaystyle =&\displaystyle [value] \end{array}\]