Question

Yea thank goodness is not the only part I know

Answer 1

$y=\sqrt{-5e^x+12x}$

a.

Insides:

$u=-5e^x+12x$

Outside:

$y=\sqrt{u}$

b.

Use the chain rule to find the derivate:

$\frac{d}{dx}(y(u))=\frac{d}{du}(y(u))*\frac{d}{dx}u$$\begin{gathered} \frac{d}{dx}(y)=\frac{d}{du}\sqrt{u}*\frac{d}{dx}(-5e^x+12x) \\ \\ \frac{dy}{dx}=\frac{1}{2\sqrt{u}}*(-5e^x+12) \\ \\ \frac{dy}{dx}=\frac{-5e^x+12}{2\sqrt{u}} \\ \\ \frac{dy}{dx}=\frac{-5e^x+12}{2\sqrt{-5e^x+12x}} \\ \end{gathered}$

Derivate:

$\frac{dy}{dx}=\frac{-5e^x+12}{2\sqrt{-5e^x+12x}}$