# Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

$x^{\cos ^{-1} x}$

Solution:

Let $y=x^{\cos ^{-1} x}$

Taking log both the sides:

$\Rightarrow \log y=\log x^{\cos ^{-1} x}$

$\Rightarrow \log y=\cos ^{-1} x \log x\left\{\log x^{a}=\operatorname{alog} x\right\}$

Differentiating with respect to $\mathrm{x}$ :

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{y})}{\mathrm{dx}}=\frac{\mathrm{d}\left(\cos ^{-1} \mathrm{x} \log \mathrm{x}\right)}{\mathrm{dx}}$

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{y})}{\mathrm{dx}}=\cos ^{-1} \mathrm{x} \times \frac{\mathrm{d}(\log \mathrm{x})}{\mathrm{dx}}+\log \mathrm{x} \times \frac{\mathrm{d}\left(\cos ^{-1} \mathrm{x}\right)}{\mathrm{dx}}$

$\left\{\right.$ Using product rule, $\left.\frac{\mathrm{d}(\mathrm{uv})}{\mathrm{dx}}=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{\cos ^{-1} x}{x}+\log x\left(\frac{-1}{\sqrt{1-x^{2}}}\right)$

$\left\{\frac{d(\log u)}{d x}=\frac{1}{u} \frac{d u}{d x} \& \frac{d\left(\cos ^{-1} x\right)}{d x}=\frac{-1}{\sqrt{1-x^{2}}}\right\}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{\cos ^{-1} x}{x}-\frac{\log x}{\sqrt{1-x^{2}}}$

$\Rightarrow \frac{d y}{d x}=y\left\{\frac{\cos ^{-1} x}{x}-\frac{\log x}{\sqrt{1-x^{2}}}\right\}$

Put the value of $y=x^{\cos ^{-1} x}:$

$\Rightarrow \frac{d y}{d x}=x^{\cos ^{-1} x}\left\{\frac{\cos ^{-1} x}{x}-\frac{\log x}{\sqrt{1-x^{2}}}\right\}$