# How do you find the derivative of sqrt(x ln(x^4))?

Mar 23, 2018

$\frac{\ln \left({x}^{4}\right) + 4}{2 \sqrt{x \ln \left({x}^{4}\right)}}$

#### Explanation:

Let's rewrite it as:
$\left[{\left(x \ln \left({x}^{4}\right)\right)}^{\frac{1}{2}}\right] '$

Now we have to derivate from the outside to the inside using the chain rule.
$\frac{1}{2} {\left[x \ln \left({x}^{4}\right)\right]}^{- \frac{1}{2}} \cdot \left[x \ln \left({x}^{4}\right)\right] '$

Here we got a derivative of a product
$\frac{1}{2} {\left(x \ln \left({x}^{4}\right)\right)}^{- \frac{1}{2}} \cdot \left[\left(x '\right) \ln \left({x}^{4}\right) + x \left(\ln \left({x}^{4}\right)\right) '\right]$

$\frac{1}{2} {\left(x \ln \left({x}^{4}\right)\right)}^{- \frac{1}{2}} \cdot \left[1 \cdot \ln \left({x}^{4}\right) + x \left(\frac{1}{x} ^ 4 \cdot 4 {x}^{3}\right)\right]$

Just using basic algebra to get a semplified version:
$\frac{1}{2} {\left(x \ln \left({x}^{4}\right)\right)}^{- \frac{1}{2}} \cdot \left[\ln \left({x}^{4}\right) + 4\right]$

And we get the solution:
$\frac{\ln \left({x}^{4}\right) + 4}{2 \sqrt{x \ln \left({x}^{4}\right)}}$

By the way you can even rewrite the inital problem to make it more simple:
$\sqrt{4 x \ln \left(x\right)}$
$\setminus \sqrt{4} \sqrt{x \ln \left(x\right)}$
$2 \sqrt{x \ln \left(x\right)}$