# How do you find the derivative of  arcsin^3(5x)?

$f ' \left(x\right) = \setminus \frac{15 {\left(\setminus {\sin}^{- 1} \left(5 x\right)\right)}^{2}}{\setminus \sqrt{1 - {\left(5 x\right)}^{2}}}$

#### Explanation:

Given function:

$f \left(x\right) = {\left(\setminus {\sin}^{- 1} \left(5 x\right)\right)}^{3}$

Differentiating above function w.r.t. $x$ using chain rule as follows

$\frac{d}{\mathrm{dx}} f \left(x\right) = \frac{d}{\mathrm{dx}} {\left(\setminus {\sin}^{- 1} \left(5 x\right)\right)}^{3}$

$f ' \left(x\right) = 3 {\left(\setminus {\sin}^{- 1} \left(5 x\right)\right)}^{2} \frac{d}{\mathrm{dx}} \left({\sin}^{- 1} \left(5 x\right)\right)$

$= 3 {\left(\setminus {\sin}^{- 1} \left(5 x\right)\right)}^{2} \setminus \frac{1}{\setminus \sqrt{1 - {\left(5 x\right)}^{2}}} \frac{d}{\mathrm{dx}} \left(5 x\right)$

$= 3 {\left(\setminus {\sin}^{- 1} \left(5 x\right)\right)}^{2} \setminus \frac{1}{\setminus \sqrt{1 - 25 {x}^{2}}} \left(5\right)$

$= \setminus \frac{15 {\left(\setminus {\sin}^{- 1} \left(5 x\right)\right)}^{2}}{\setminus \sqrt{1 - {\left(5 x\right)}^{2}}}$