Chapter 3 - Derivatives - 3.7 The Chain Rule - 3.7 Exercises - Page 192: 79

\[\begin{align} & \mathbf{a}.\text{ }h\left( 4 \right)=9\text{ and }h'\left( 4 \right)=-6 \\ & \mathbf{b}.y=-6x+33 \\ \end{align}\]

\[\begin{align} & \text{Let }h\left( x \right)=f\left( g\left( x \right) \right) \\ & \\ & \mathbf{a}\mathbf{.}\text{ Calculate }h\left( 4 \right)\text{ and }h'\left( 4 \right) \\ & h\left( 4 \right)=f\left( g\left( 4 \right) \right) \\ & \text{The graph }g\text{ passes through the point }\left( 4,7 \right)\to g\left( 4 \right)=7,\text{ then} \\ & h\left( 4 \right)=f\left( 7 \right) \\ & \text{The graph }f\text{ passes through the point }\left( 7,9 \right)\to f\left( 7 \right)=9,\text{ then} \\ & h\left( 4 \right)=9 \\ & and \\ & h'\left( x \right)=\frac{d}{dx}\left[ f\left( g\left( x \right) \right) \right] \\ & h'\left( 4 \right)=f'\left( g\left( 4 \right) \right)g'\left( 4 \right) \\ & \text{The tangent line to the graph of }f\text{ at }\left( 4,7 \right)\text{ is }y=-2x+23,\text{ recall} \\ & \text{that }\underbrace{y=-2x+23}_{y=mx+b}\Rightarrow m=-2,\text{ and the slope of the tangent line at} \\ & \text{the point }\left( x,y \right)\text{ is }m=f'\left( x \right),\text{ then }f'\left( g\left( 4 \right) \right)=-2 \\ & and \\ & \text{The tangent line to the graph of }g\text{ at }\left( 7,9 \right)\text{ is }y=3x-5,\text{ recall} \\ & \text{that }\underbrace{y=3x-5}_{y=mx+b}\Rightarrow m=3,\text{ and the slope of the tangent line at} \\ & \text{the point }\left( x,y \right)\text{ is }m=f'\left( x \right),\text{ then }f'\left( g\left( 4 \right) \right)=3,\text{ then} \\ & h'\left( 4 \right)=f'\left( g\left( 4 \right) \right)g'\left( 4 \right) \\ & h'\left( 4 \right)=\left( -2 \right)\left( 3 \right) \\ & h'\left( 4 \right)=-6 \\ & \\ & \mathbf{b}\mathbf{.}\text{ The equation of the tangent to the tangent line to }h\text{, }x=4 \\ & \text{The point is }\left( \underbrace{4}_{{{x}_{1}}},\underbrace{h\left( 4 \right)}_{{{y}_{1}}} \right)\text{ with the slope }m=h'\left( 4 \right) \\ & y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\ & y-h\left( 4 \right)=h'\left( 4 \right)\left( x-4 \right) \\ & \text{Where }h\left( 4 \right)=9\text{ and }h'\left( 4 \right)=-6 \\ & y-9=-6\left( x-4 \right) \\ & y-9=-6x+24 \\ & \text{ }y=-6x+33 \\ \end{align}\]