# How do you differentiate  y= sqrt( (x^2 + 4x + 1)^2+2x) using the chain rule?

Dec 30, 2015

y'=(2(x^2+4x+1)(2x+4)+2)/(2sqrt((x^2+4x+1)^2+2x)

#### Explanation:

According to the chain rule:

$\frac{d}{\mathrm{dx}} \left({u}^{\frac{1}{2}}\right) = \frac{1}{2} \left({u}^{- \frac{1}{2}}\right) \cdot u ' = \frac{u '}{2 \sqrt{u}}$

Set $u = {\left({x}^{2} + 4 x + 1\right)}^{2} + 2 x$.

Using the aforementioned rule:

y'=(d/dx((x^2+4x+1)^2+2x))/(2sqrt((x^2+4x+1)^2+2x)

Find $\frac{d}{\mathrm{dx}} \left({\left({x}^{2} + 4 x + 1\right)}^{2} + 2 x\right)$.

For the first term, use chain rule again.

$\implies 2 \left({x}^{2} + 4 x + 1\right) \left(2 x + 4\right) + 2$

y'=(2(x^2+4x+1)(2x+4)+2)/(2sqrt((x^2+4x+1)^2+2x)