Derivative Rules Question 3

The figure above shows the graph of the function $g$ and the line tangent to the graph of $g$ at $x=-1$ . Let $h$ be the function given by $h(x)=e^x\cdot g(x)$ . What is the value of $h'(-1)$ ?

\hskip{.8em}\dfrac{9}{e}

-\dfrac{3}{e}

-\dfrac{6}{e}-\dfrac{3}{e^2}

-6

Approach

h(x)=e^x\cdot g(x)

h'(x)=e^xg'(x)+g(x)e^x

h'(x)=e^x(g'(x)+g(x))

Use the graph to find $g'(x)$ and $g(x)$ .

The derivative at $g'(-1)$ is equal to the slope of the tangent line.

m =\frac{-3-3}{0-(-1)}

m = -\frac{6}{1}

g'(-1) = -6

h'(-1)=e^{-1}(g'(-1)+g(-1))

h'(-1)=\frac{1}{e}(-6+3)

= -\frac{3}{e}