Question

differentiate the function.
y = \sqrt{x^{2}+11}
\frac{dy}{dx}=\square

Explanation:

Step1: Rewrite the function

Rewrite $y = \sqrt{x^{2}+11}$ as $y=(x^{2}+11)^{\frac{1}{2}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = f(g(x))$, then $y^\prime=f^\prime(g(x))\cdot g^\prime(x)$. Let $u = x^{2}+11$, so $y = u^{\frac{1}{2}}$. First, find $\frac{dy}{du}$ and $\frac{du}{dx}$.
$\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dx}=2x$.

Step3: Substitute and simplify

$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=\frac{1}{2}(x^{2}+11)^{-\frac{1}{2}}\cdot2x=\frac{x}{\sqrt{x^{2}+11}}$.

Answer:

$\frac{x}{\sqrt{x^{2}+11}}$