Question

Differentiate ${\cos }^{-1}(4x^3-3x)$ w.r.t $x$.

• A. $\frac{3}{\sqrt{1-x^2}}$
• B. $-\frac{3}{\sqrt{1-x^2}}$
• C. $\frac{1}{\sqrt{1-x^2}}$
• D. $-\frac{1}{\sqrt{1-x}}$

Answer 1

Answer: (B)

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

(Let the given function be $y=f\left(x\right)$.

$f\left(x\right)={\cos }^{-1}(4x^3-3x)$

From the definition of first principle of differentiation,

$\frac{dy}{dx}={lim}_{x\to 0}\frac{f(x+\delta x)-f\left(x\right)}{\delta x}$

Now coming to the problem,

we know that ${\cos }^{-1}(4x^3-3x)=3{\cos }^{-1}\left(x\right)$

$\frac{dy}{dx}={lim}_{\delta x\to 0}\frac{f(x+\delta x)-f\left(x\right)}{\delta x}$

$\frac{dy}{dx}={lim}_{\delta x\to 0}\frac{3{\cos }^{-1}(x+\delta x)-3{\cos }^{-1}x}{\delta x}$

$\frac{dy}{dx}=3{lim}_{\delta x\to 0}\frac{{\cos }^{-1}(x+\delta x)-{\cos }^{-1}\left(x\right)}{\delta x}$

We know derivative of ${]cos}^{-1}x$ is $\frac{-1}{\sqrt{1-x^2}}$

Therefore, $\frac{dy}{dx}=-\frac{3}{\sqrt{1-x^2}}$.