Question

If $y=log{x}^3+3si{n}^{-1}x+k{x}^2$, then find $\frac{dy}{dx}$

• A. $3\cdot \frac{1}{x}+3\cdot \frac{1}{\sqrt{1-x^2}}+k\left(2x\right)$
• B. $3\cdot \frac{1}{x^3}+3\cdot \frac{1}{\sqrt{1-x^2}}+k\left(2x\right)$
• C. $3\cdot \frac{1}{x}-3\cdot \frac{1}{\sqrt{1-x^2}}+k\left(2x\right)$
• D. $3\cdot \frac{1}{x}+3\cdot \frac{1}{\sqrt{1-x^2}}+2x$

Answer 1

Answer: (A)

$3\cdot \frac{1}{x}+3\cdot \frac{1}{\sqrt{1-x^2}}+k\left(2x\right)$

Here, $y=\log x^3+3{\sin }^{-1}x+k{x}^2$

On differentiating we get

$\frac{dy}{dx}=\frac{d}{dx}\left[\log x^3\right]+\frac{d}{dx}\left[3{\sin }^{-1}x\right]+\frac{d}{dx}\left[k{x}^2\right]$

$=3\frac{d}{dx}\left[\log x\right]+3\frac{d}{dx}\left({\sin }^{-1}x\right)+k\frac{d}{dx}\left(x^2\right)$

$=3\cdot \frac{1}{x}+3\cdot \frac{1}{\sqrt{1-x^2}}+k\left(2x\right)$