Question

If $y={\cos }^{-1}\left(\ln x\right)$, then value of $\frac{dy}{dx}$ is:

• A. $-\frac{1}{x\sqrt{1+\ln x}}$
• B. $\frac{1}{x\sqrt{1+(\ln x{)}^2}}$
• C. $-\frac{1}{x\sqrt{1-(\ln x{)}^2}}$
• D. None of these

Answer 1

Answer: (C)

$-\frac{1}{x\sqrt{1-(\ln x{)}^2}}$

Given $y={\cos }^{-1}\left(\ln x\right)$

To find $\frac{dy}{dx}$,

$\cos y=\ln x$

Differentiating with respect to $x$ on both sides gives,

$-\sin y\frac{dy}{dx}=\frac{1}{x}$

$⟹\frac{dy}{dx}=\frac{-1}{x\sin y}$

We know that ${\left(\sin x\right)}^2+{\left(\cos x\right)}^2=1$

$⟹\sin y=\sqrt{1-{\left(\ln x\right)}^2}$

$\therefore \frac{dy}{dx}=-\frac{1}{x\left(\sqrt{1-{\left(\ln x\right)}^2}\right)}$