What is the derivative of this function sin^-1 (5x)?

Apr 13, 2017

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

Explanation:

Recall:

$\left({\sin}^{- 1} u\right) ' = \frac{1}{\sqrt{1 - {u}^{2}}}$

By the above rule ($u = 5 x$) along with Chain Rule,

(sin^(-1)(5x))'=1/sqrt(1-(5x)^2)cdot(5x)' =5/sqrt(1-25x^2)

I hope that this was clear.

Apr 13, 2017

$\frac{5}{\sqrt{1 - 25 {x}^{2}}}$

Explanation:

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \frac{d}{\mathrm{dx}} \left({\sin}^{-} 1 x\right) = \frac{1}{\sqrt{1 - {x}^{2}}}$

"and " d/dx(sin^-1(f(x)))=(f'(x))/(sqrt(1-f(x)^2)

$\Rightarrow \frac{d}{\mathrm{dx}} \left({\sin}^{-} 1 \left(5 x\right)\right)$

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

$= \frac{5}{\sqrt{1 - 25 {x}^{2}}}$