# How do you use the chain rule to differentiate y=(x^2+1)^(1/2)?

Feb 26, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{\sqrt{{x}^{2} + 1}}$

#### Explanation:

$y = {\left({x}^{2} + 1\right)}^{\frac{1}{2}}$

Apply the power rule and the chain rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \cdot {\left({x}^{2} + 1\right)}^{- \frac{1}{2}} \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right)$

Apply the power rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \cdot {\left({x}^{2} + 1\right)}^{- \frac{1}{2}} \cdot \left(2 x + 0\right)$

= (cancel2x)/(cancel2* sqrt(x^2+1)

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

Feb 26, 2017

Recall that the chain rule is similar as the power rule but has one more step.

#### Explanation:

The chain rule states that the derivative of $f \left(g \left(x\right)\right)$ is $f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$.

In this instance,

$f \left(x\right) = g {\left(x\right)}^{\frac{1}{2}}$

$g \left(x\right) = {x}^{2} + 1$

So, the derivative of the composition $f \left(g \left(x\right)\right)$ is $f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$, which is:

$\left(\frac{1}{2}\right) \cdot {\left({x}^{2} + 1\right)}^{- \frac{1}{2}} \cdot 2 x$

Simplified, we get:

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