Question

If $y=x^2+si{n}^{-1}x+lo{g}_{e}x$, find $\frac{dy}{dx}$

• A. $\frac{dy}{dx}=2x+\frac{1}{\sqrt{1-x^2}}+\frac{1}{x}$
• B. $\frac{dy}{dx}=x+\frac{1}{\sqrt{1-x^2}}+\frac{1}{x}$
• C. $\frac{dy}{dx}=2x+\frac{1}{\sqrt{1-x^2}}-\frac{1}{x}$
• D. $\frac{dy}{dx}=2x-\frac{1}{\sqrt{1-x^2}}+\frac{1}{x}$

Answer 1

The correct option is A $\frac{dy}{dx}=2x+\frac{1}{\sqrt{1-x^2}}+\frac{1}{x}$

$y=x^2+si{n}^{-1}+lo{g}_{e}x$
On differentiating, we get
$\frac{dy}{dx}=\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}si{n}^{-1}x)+\frac{d}{dx}(lo{g}_{e}x)$
or $\frac{dy}{dx}=2(x{)}^{2-1}+\frac{1}{\sqrt{1-x^2}}+\frac{d}{dx}(lo{g}_{e}x)$
$\frac{dy}{dx}=2x+\frac{1}{\sqrt{1-x^2}}+\frac{1}{x}$