Suppose $h=f\circ g$. Find $h^\prime (1)$ given that: $$f(1) =2 \quad f^\prime (1) =1 \quad f^\prime (2) =-4 \quad g(1) =2 \quad g^\prime (1) =-3\quad g^\prime (2) =5$$

Solution Based on the Chain Rule using prime notation, we have $$h^\prime ( x_{0}) =( f\circ g)^\prime ( x_{0}) =f^\prime (g( x_{0})) g^\prime ( x_{0})$$

When $x_{0}=1,$ $$\begin{eqnarray*} && h^\prime (1) =f^\prime (g(1)) g^\prime(1) \underset{\underset{\color{#0066A7}{{g(1)=2; g^\prime (1)=-3}}}{\color{#0066A7}{{\uparrow}}}}{=} f^\prime (2) \cdot ( -3) \underset{\underset{\color{#0066A7}{{f^\prime (2) =-4}}} {\color{#0066A7}{{\uparrow}}}}{=}(-4) ( -3) =12 \\ \end{eqnarray*}$$