$$\begin{array}{|c||c|c|c|c|} \hline x & f(x) & f'(x) & g(x) & g'(x) \\ \hline 1 & 3 & -2 & -3 & 4 \\ \hline \end{array}$$

The table above gives the values of the differentiable functions $f$ and $g$ and their derivatives at $x=1$. If $h(x)=(2f(x)+3)(1+g(x))$, then $h'(1)=$

Summary Submit

$$h(x)=(2f(x)+3)(1+g(x))$$

Using the product rule:

$$h'(x)=(2f(x)+3)(g'(x))+(1+g(x))(2f'(x))$$ $$h'(1)=(2f(1)+3)(g'(1))+(1+g(1))(2f'(1))$$ $$(2(3)+3)(4)+(1+(-3))(2(-2))$$ $$(9)(4)+(-2)(-4)$$ $$36+8$$ $$=\boxed{44}$$