Question

Find the derivative of $y$ with respect to $x$ at $x=1$, where function $y$ is expressed as $y=\sqrt{x^3+1}$ .

• A. $\frac{1}{2\sqrt{2}}$
  
• B. $\frac{3}{\sqrt{2}}$
• C. $\frac{3\sqrt{2}}{\sqrt{5}}$
• D. $\frac{3}{2\sqrt{2}}$
  

Answer 1

Answer: (D)

$\frac{3}{2\sqrt{2}}$


We have, $y=\sqrt{x^3+1}$

let $\mu \left(x\right)=x^3+1$

$\therefore y=\sqrt{\mu },\frac{dy}{d\mu }=\frac{1}{2\sqrt{\mu }}$

and we know from chain rule:

$\frac{dy}{dx}=\frac{dy}{d\mu }.\frac{d\mu }{dx}$

and $\frac{d\mu }{dx}=\frac{d}{dx}(x^3+1)=3x^2$

$\Rightarrow \frac{dy}{dx}=\frac{3x^2}{2\sqrt{x^3+1}}$

At $(x=1)$

$\Rightarrow \frac{dy}{dx}=\frac{3\times 1^2}{2\sqrt{1^3+1}}=\frac{3}{2\sqrt{2}}$