Question

find the derivative.
$\frac{dy}{dx}$ if $y(x)=sqrt{x^{2}+1}$
$\frac{dy}{dx}=square$

Explanation:

Step1: Rewrite the function

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

Step2: Apply the chain - rule

The chain - rule states that if $y = u^{\frac{1}{2}}$ and $u=x^{2}+1$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$: $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$. Second, find $\frac{du}{dx}$: $\frac{du}{dx}=2x$.

Step3: Substitute and simplify

Substitute $u = x^{2}+1$ back into $\frac{dy}{du}$ and multiply by $\frac{du}{dx}$: $\frac{dy}{dx}=\frac{1}{2}(x^{2}+1)^{-\frac{1}{2}}\cdot2x=\frac{x}{\sqrt{x^{2}+1}}$.

Answer:

$\frac{x}{\sqrt{x^{2}+1}}$