Question

differentiate.
y = \sqrt{e^{x}+15}
\frac{dy}{dx}=\square

Explanation:

Step1: Rewrite the function

Let $u = e^{x}+15$, then $y=\sqrt{u}=u^{\frac{1}{2}}$.

Step2: Differentiate $y$ with respect to $u$

Using the power - rule $\frac{d}{du}(u^{n})=nu^{n - 1}$, we have $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{u}}$.

Step3: Differentiate $u$ with respect to $x$

Since $\frac{d}{dx}(e^{x}) = e^{x}$ and $\frac{d}{dx}(15)=0$, then $\frac{du}{dx}=e^{x}$.

Step4: Use the chain - rule

The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substituting $\frac{dy}{du}=\frac{1}{2\sqrt{u}}$ and $\frac{du}{dx}=e^{x}$ into the chain - rule formula, and replacing $u = e^{x}+15$, we get $\frac{dy}{dx}=\frac{e^{x}}{2\sqrt{e^{x}+15}}$.

Answer:

$\frac{e^{x}}{2\sqrt{e^{x}+15}}$