Title: CR_Diff_06
Course: Demonstrating STACK (v4.4.7)

Chain rule for differentiation

Differentiate [value] with respect to [value].
Feedback: In order to solve this problem we will have to use the chain rule calc_chain_rule as follows: Define \(f([value])=[value]\) and \(g([value])=[value]\). So \(f([value])=g([value])^[value]\). So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}[[value]] &\displaystyle =&\displaystyle \frac{df}{dg}\cdot \frac{dg}{d[value]} \\ &\displaystyle =&\displaystyle \frac{d}{dg}(g^[value])\cdot \frac{d}{d[value]}([value]) \\ &\displaystyle =&\displaystyle [value]g^{[value]-1}\cdot [value] \\ &\displaystyle =&\displaystyle [value]([value])^[value] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Differentiate [value] with respect to [value].
Feedback: In order to solve this problem we will have to use the chain rule calc_chain_rule as follows: Define \(f([value])=[value]\) and \(g([value])=[value]\). So \(f([value])=g([value])^[value]\). So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}[[value]] &\displaystyle =&\displaystyle \frac{df}{dg}\cdot \frac{dg}{d[value]} \\ &\displaystyle =&\displaystyle \frac{d}{dg}(g^[value])\cdot \frac{d}{d[value]}([value]) \\ &\displaystyle =&\displaystyle [value]g^{[value]-1}\cdot (-[value])[value]^{[value]-1} \\ &\displaystyle =&\displaystyle -[value][value]^[value]([value])^[value] \\ &\displaystyle =&\displaystyle -[value] \end{array}\]

Differentiate \[[value]\] with respect to [value].
Feedback: In order to solve this problem we will have to use the chain rule calc_chain_rule as follows: Define \(f([value])=[value]\) and \(g([value])=[value]\). So \(f([value])=g([value])^[value]\). So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}[[value]] &\displaystyle =&\displaystyle \frac{df}{dg}\cdot \frac{dg}{d[value]} \\ &\displaystyle =&\displaystyle \frac{d}{dg}(g^[value])\cdot \frac{d}{d[value]}([value]) \\ &\displaystyle =&\displaystyle [value]g^{[value]-1}\cdot [value] \\ &\displaystyle =&\displaystyle [value]([value])^[value] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Differentiate [value] with respect to [value].
Feedback: In order to solve this problem we will have to use the chain rule calc_chain_rule as follows: Define \(f([value])=[value]\) and \(g([value])=[value]\). So \(f([value])=g([value])^[value]\). So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}[[value]] &\displaystyle =&\displaystyle \frac{df}{dg}\cdot \frac{dg}{d[value]} \\ &\displaystyle =&\displaystyle \frac{d}{dg}(g^[value])\cdot \frac{d}{d[value]}([value]) \\ &\displaystyle =&\displaystyle [value]g^{[value]-1}\cdot [value] \\ &\displaystyle =&\displaystyle [value]([value])^[value][[value]] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Differentiate \[[value]\] with respect to [value].
Feedback: In order to solve this problem we will have to use the chain rule calc_chain_rule as follows: Define \(f([value])=[value]\) and \(g([value])=[value]\). So \(f([value])=g([value])^[value]\). So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}[[value]] &\displaystyle =&\displaystyle \frac{df}{dg}\cdot \frac{dg}{d[value]} \\ &\displaystyle =&\displaystyle \frac{d}{dg}(g^[value])\cdot \frac{d}{d[value]}([value]) \\ &\displaystyle =&\displaystyle [value]g^{[value]-1}\cdot [value] \\ &\displaystyle =&\displaystyle [value]([value])^[value][[value]] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Differentiate [value] with respect to [value].
Feedback: In order to solve this problem we will have to use the chain rule calc_chain_rule as follows: Define \(f([value])=[value]\) and \(g([value])=[value]\). So \(f([value])=sin(g([value]))\). So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}[[value]] &\displaystyle =&\displaystyle \frac{df}{dg}\cdot \frac{dg}{d[value]} \\ &\displaystyle =&\displaystyle \frac{d}{dg}[sin(g)]\cdot \frac{d}{d[value]}([value]) \\ &\displaystyle =&\displaystyle [value]\cdot [value] \\ &\displaystyle =&\displaystyle [[value]][[value]] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Find the following derivative: \[ \frac{d}{d[value]}[[value]]. \]
Feedback: In order to solve this problem we will have to use the chain rule calc_chain_rule as follows: Define \(f([value]):=[value]\) and \(g([value]):=[value]\). Then \(f([value])=sin[g([value])]\). So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}[[value]] &\displaystyle =&\displaystyle \frac{df}{dg}\cdot \frac{dg}{d[value]} \\ &\displaystyle =&\displaystyle \frac{d}{dg}[sin(g)]\cdot \frac{d}{d[value]}[[value]] \\ &\displaystyle =&\displaystyle [value]\cdot [[value]] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Differentiate [value] with respect to [value].
Feedback: In order to solve this problem we will have to use the chain rule calc_chain_rule as follows: Define \(f([value])=[value]\) and \(g([value])=[value]\). So \(f([value])=(g([value]))^[value]\). So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}[[value]] &\displaystyle =&\displaystyle \frac{df}{dg}\cdot \frac{dg}{d[value]} \\ &\displaystyle =&\displaystyle \frac{d}{dg}[tan(g)]\cdot \frac{d}{d[value]}([value]) \\ &\displaystyle =&\displaystyle [value]\cdot [value] \\ &\displaystyle =&\displaystyle [[value]][[value]] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Differentiate [value] with respect to [value].
Feedback: In order to solve this problem we will have to use the chain rule calc_chain_rule as follows: Define \(f([value])=[value]\) and \(g([value])=[value]\). So \(f([value])=(g([value]))^[value]\). So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}[[value]] &\displaystyle =&\displaystyle \frac{df}{dg}\cdot \frac{dg}{d[value]} \\ &\displaystyle =&\displaystyle \frac{d}{dg}[(g)^[value]]\cdot \frac{d}{d[value]}([value]) \\ &\displaystyle =&\displaystyle [value]\cdot [value] \\ &\displaystyle =&\displaystyle [[value]][[value]] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Differentiate \[[value]\] with respect to [value].
Feedback: In order to solve this problem we will have to use the chain rule calc_chain_rule as follows: Define \(f([value])=[value]\) and \(g([value])=[value]\). So \(f([value])=(g([value]))^[value]\). So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}\left[[value]\right] &\displaystyle =&\displaystyle \frac{df}{dg}\cdot \frac{dg}{d[value]} \\ &\displaystyle =&\displaystyle \frac{d}{dg}[(g)^[value]]\cdot \frac{d}{d[value]}([value]) \\ &\displaystyle =&\displaystyle [value]\cdot [value] \\ &\displaystyle =&\displaystyle \left[[value]\right][[value]] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Differentiate [value] with respect to [value].
Feedback: In order to solve this problem we will have to use the chain rule calc_chain_rule as follows: Define \(f([value])=[value]\) and \(g([value])=[value]\). So \[f([value])=e^{g([value])}\]. So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}[[value]] &\displaystyle =&\displaystyle \frac{df}{dg}\cdot \frac{dg}{d[value]} \\ &\displaystyle =&\displaystyle \frac{d}{dg}[e^{g}]\cdot \frac{d}{d[value]}([value]) \\ &\displaystyle =&\displaystyle [value]\cdot [value] \\ &\displaystyle =&\displaystyle \left[[value]\right][[value]] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Differentiate [value] with respect to [value].
Feedback: In order to solve this problem we will have to use the chain rule calc_chain_rule as follows: Define \(f([value])=[value]\) and \(g([value])=[value]\). So \[f([value])=e^{(g([value]))}\]. So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}[[value]] &\displaystyle =&\displaystyle \frac{df}{dg}\cdot \frac{dg}{d[value]} \\ &\displaystyle =&\displaystyle \frac{d}{dg}[e^{g}]\cdot \frac{d}{d[value]}([value]) \\ &\displaystyle =&\displaystyle [value]\cdot [value] \\ &\displaystyle =&\displaystyle [[value]][[value]] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Differentiate [value] with respect to [value].
Feedback: In order to solve this problem we will have to use the chain rule calc_chain_rule as follows: Define \(f([value])=[value]\) and \(g([value])=[value]\). So \(f([value])=ln(g([value]))\). So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}[[value]] &\displaystyle =&\displaystyle \frac{df}{dg}\cdot \frac{dg}{d[value]} \\ &\displaystyle =&\displaystyle \frac{d}{dg}[ln(g)]\cdot \frac{d}{d[value]}([value]) \\ &\displaystyle =&\displaystyle [value]\cdot [value] \\ &\displaystyle =&\displaystyle \left[[value]\right][[value]] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Find the following derivative:
\[ \frac{d}{d[value]}[[value]]. \]
Feedback: In order to solve this problem we will have to use the chain rule calc_chain_rule as follows: Define \(f([value]):=[value]\) and \(g([value]):=[value]\). Then \(f([value])=sin[g([value])]\). So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}[[value]] &\displaystyle =&\displaystyle \frac{df}{dg}\cdot \frac{dg}{d[value]} \\ &\displaystyle =&\displaystyle \frac{d}{dg}[sin(g)]\cdot \frac{d}{d[value]}[[value]] \\ &\displaystyle =&\displaystyle [value]\cdot \frac[value]{[value] [value]} \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Differentiate [value] with respect to [value].
Feedback: In order to solve this problem we will have to use the chain rule calc_chain_rule as follows: Define \(f([value])=[value]\) and \(g([value])=[value]\). So \(f([value])=cos(g([value]))\). So, \[\begin{array}{rcl}\displaystyle \frac{d}{d[value]}[[value]] &\displaystyle =&\displaystyle \frac{df}{dg}\cdot \frac{dg}{d[value]} \\ &\displaystyle =&\displaystyle \frac{d}{dg}[cos(g)]\cdot \frac{d}{d[value]}([value]) \\ &\displaystyle =&\displaystyle [value]\cdot [value] \\ &\displaystyle =&\displaystyle [[value]][[value]] \\ &\displaystyle =&\displaystyle [value] \end{array}\]