Question

Differentiate sin inverse(3x-4x^3)

Answer 1

here is your answer.......

Answer 2

Thus, The answer is $\frac{3-12x^{2} }{\sqrt{1-9x^{2} +24x^{4}-16x^{6}} }$

Step-by-step explanation:

Given,

Differentiate $sin^{-1}(3x-4x^{3} )$

Differentiate above equation with respect to $x$,

∴$\frac{d}{dx}sin^{-1}(3x-4x^{3} )$

$=\frac{1}{\sqrt{1-(3x-4x^{3})^{2}} } \frac{d}{dx}(3x-4x^{3} )$  (∵ $\frac{d}{dx} sin^{-1} x=\frac{1}{\sqrt{1-x^{2} } }$)

$=\frac{1}{\sqrt{1-(9x^{2} -2\times 3x\times 4x^{3}+16x^{6} ) } } (3-12x^{2} )$

$=\frac{3-12x^{2} }{\sqrt{1-9x^{2} +24x^{4}-16x^{6}} }$

So, The answer is $\frac{3-12x^{2} }{\sqrt{1-9x^{2} +24x^{4}-16x^{6}} }$