Question

Differentiate the following w.r.t. x : cos^–1(3x – 4x³)

Answer 1

Let y = cos^–1(3x – 4x³)
Put x = sinθ.
Then θ = sin^–1x
∴ y = cos^–1(3sinθ - 4sin³θ)
= cos^–1(sin3θ)

= ${\displaystyle {\cos }^{-1}\left[\cos (\frac{\pi }{2}-3\theta )\right]}$

= ${\displaystyle \frac{\pi }{2}-3\theta }$

= ${\displaystyle \frac{\pi }{2}-3{\sin }^{-1}x}$
Differentiating w.r.t. x, we get
${\displaystyle \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\frac{\pi }{2}-3{\sin }^{-1}x)}$

= ${\displaystyle \frac{\text{dy}}{\text{dx}}\left(\frac{\pi }{2}\right)-3\frac{\text{d}}{\text{dx}}\left({\sin }^{-1}x\right)}$

= ${\displaystyle 0-3\times \frac{1}{\sqrt{1-x^2}}}$

= ${\displaystyle \frac{-3}{\sqrt{1-x^2}}}$.