?Find equations of the tangent line and normal line to the curve at the given point.y =

Step 1 of 3

Differentiate

\(\begin{array}{l} \frac{d y}{d x}=\frac{d\left(x^{4}+2 e^{x}\right)}{d x} \\ \frac{d y}{d x}=\frac{d\left(x^{4}\right)}{d x}+\frac{d\left(2 e^{x}\right)}{d x} \\ \frac{d y}{d x}=4 x^{4-1}+2 e^{x} \\ \frac{d y}{d x}=4 x^{3}+2 e^{x} \end{array}\)

Slope of tangent at \((0,2)=\left.\frac{d y}{d x}\right|_{x=0}=4 \cdot 0^{3}+2 e^{0}=0+2=2\)

Product of slopes of perpendicular lines is \(-1\)

Therefore slope of the normal line should be \(-\frac{1}{2}\)