# How do you find the derivative of f(x) = arcsin (2x + 5)?

Sep 15, 2016

$\frac{2}{\sqrt{4 {x}^{2} + 20 x + 24} \cdot i}$

#### Explanation:

We have: $f \left(x\right) = \arcsin \left(2 x + 5\right)$

This function can be differentiated using the "chain rule".

Let $u = 2 x + 5 \implies u ' = 2$ and $v = \arcsin \left(u\right) \implies v ' = \frac{1}{\sqrt{1 - {u}^{2}}}$:

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

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

We can now replace $u$ with $2 x + 5$:

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

=> (d) / (dx) (arcsin(2 x + 5)) = (2) / (sqrt(1 - (4 x^(2) + 20 x + 25))

$\implies \frac{d}{\mathrm{dx}} \left(\arcsin \left(2 x + 5\right)\right) = \frac{2}{\sqrt{- 4 {x}^{2} - 20 x - 24}}$

$\implies \frac{d}{\mathrm{dx}} \left(\arcsin \left(2 x + 5\right)\right) = \frac{2}{\sqrt{- \left(4 {x}^{2} + 20 x + 24\right)}}$

$\implies \frac{d}{\mathrm{dx}} \left(\arcsin \left(2 x + 5\right)\right) = \frac{2}{\sqrt{4 {x}^{2} + 20 x + 24} \cdot i}$