differentiate the function.\ny = \\sqrt{x^{2}+12}\n\\frac{dy}{dx}=\\square

Explanation:

Step1: Rewrite the function

Rewrite $y = \sqrt{x^{2}+12}$ as $y=(x^{2}+12)^{\frac{1}{2}}$.

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 $u = x^{2}+12$, so $y = u^{\frac{1}{2}}$. First, find $\frac{dy}{du}$ and $\frac{du}{dx}$. $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ (using the power rule $\frac{d}{du}(u^{n})=nu^{n - 1}$ with $n=\frac{1}{2}$), and $\frac{du}{dx}=2x$.

Step3: Calculate $\frac{dy}{dx}$

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dx}=2x$ into the formula. Replace $u$ with $x^{2}+12$. $\frac{dy}{dx}=\frac{1}{2}(x^{2}+12)^{-\frac{1}{2}}\cdot2x$. Simplify the expression: $\frac{dy}{dx}=\frac{x}{\sqrt{x^{2}+12}}$.

Answer:

$\frac{x}{\sqrt{x^{2}+12}}$