# How do you find the derivative of arcsin^5(4x+4) ?

Mar 31, 2018

$\frac{d}{\mathrm{dx}} \left({\arcsin}^{5} \left(4 x + 4\right)\right) = \frac{20 {\arcsin}^{4} \left(4 x + 4\right)}{\sqrt{1 - {\left(4 x + 4\right)}^{2}}}$

#### Explanation:

Let's take this derivative step by step.

First, we recognize that we need an order to take these in.
We should begin with the power rule by thinking of
${\arcsin}^{5} \left(4 x + 4\right) = {\left(\arcsin \left(4 x + 4\right)\right)}^{5}$

So we already have
$\frac{d}{\mathrm{dx}} \left({\arcsin}^{5} \left(4 x + 4\right)\right) = \frac{d}{\mathrm{dx}} \left({\left(\arcsin \left(4 x + 4\right)\right)}^{5}\right)$
$= 5 \arcsin {\left(4 x + 4\right)}^{4} \setminus \frac{d}{\mathrm{dx}} \left(\arcsin \left(4 x + 4\right)\right)$

Now we can use the derivative of arcsine to complete the derivation:

$= 5 \arcsin {\left(4 x + 4\right)}^{4} \cdot \frac{1}{\sqrt{1 - {\left(4 x + 4\right)}^{2}}} \cdot \frac{d}{\mathrm{dx}} \left(4 x + 4\right)$
$= \frac{20 {\arcsin}^{4} \left(4 x + 4\right)}{\sqrt{1 - {\left(4 x + 4\right)}^{2}}}$