Question

If $y=\frac{\sqrt{x}(2x+3{)}^2}{\sqrt{x+1}}$, then $\frac{dy}{dx}$ is equal to

• A. $y[\frac{1}{2x}+\frac{4}{2x+3}-\frac{1}{2(x+1)}]$
• B. $y[\frac{1}{3x}+\frac{4}{2x+3}+\frac{1}{2(x+1)}]$
• C. $y[\frac{1}{3x}+\frac{4}{2x+3}+\frac{1}{x+1}]$
• D. $Noneofthese$

Answer 1

The correct option is A $y[\frac{1}{2x}+\frac{4}{2x+3}-\frac{1}{2(x+1)}]$

$y=\frac{\sqrt{x}{(2x+3)}^2}{\sqrt{x+1}}$

Applying $\log$ on both sides, we get

$\log y=\log \left[\frac{\sqrt{x}{(2x+3)}^2}{\sqrt{x+1}}\right]$

Using the identity $\log \frac{a}{b}=\log a-\log b$ we have

$\log y=\log \left[\sqrt{x}{(2x+3)}^2\right]-\log \sqrt{x+1}$

Using the identity $\log ab=\log a+\log b$ we have

$\log y=\log \sqrt{x}+\log {(2x+3)}^2-\log \sqrt{x+1}$

Using the identity $\log a^m=m\log a$ we have

$\log y=\log \sqrt{x}+2\log (2x+3)-\log \sqrt{x+1}$

Differentiating both sides w.r.t $x$ we get

$\frac{1}{y}\frac{dy}{dx}=\frac{1}{\sqrt{x}}\times \frac{d}{dx}\left(\sqrt{x}\right)+2\frac{1}{2x+3}\times \frac{d}{dx}(2x+3)-\frac{1}{\sqrt{x+1}}\times \frac{d}{dx}\left(\sqrt{x+1}\right)$

$\frac{1}{y}\frac{dy}{dx}=\frac{1}{\sqrt{x}}\times \frac{1}{2\sqrt{x}}+2\frac{1}{2x+3}\times 2-\frac{1}{\sqrt{x+1}}\times \frac{1}{2\sqrt{x+1}}$

$\frac{1}{y}\frac{dy}{dx}=\frac{1}{2x}+\frac{4}{2x+3}-\frac{1}{2(x+1)}$

$\Rightarrow \frac{dy}{dx}=y[\frac{1}{2x}+\frac{4}{2x+3}-\frac{1}{2(x+1)}]$