Question

The derivative of $\frac{\sqrt{x}{(x+4)}^{\frac{3}{2}}}{{(4x-3)}^{\frac{4}{3}}}$ w.r.t $x$ is

• A. $\frac{\sqrt{x}{(x+4)}^{\frac{3}{2}}}{{(4x-3)}^{\frac{4}{3}}}\{\frac{1}{2x}+\frac{3}{2(x+4)}-\frac{16}{3(4x-3)}\}$
• B. $\frac{\sqrt{x}{(x+4)}^{\frac{4}{3}}}{{(4x-3)}^{\frac{3}{2}}}\{\frac{1}{2x}+\frac{3}{2(x+4)}-\frac{16}{3(4x-3)}\}$
• C. $\frac{\sqrt{x}{(x+4)}^{\frac{5}{4}}}{{(4x-3)}^{\frac{3}{4}}}\{\frac{1}{2x}+\frac{3}{2(x+4)}-\frac{16}{3(4x-3)}\}$
• D. None of these

Answer 1

The correct option is A $\frac{\sqrt{x}{(x+4)}^{\frac{3}{2}}}{{(4x-3)}^{\frac{4}{3}}}\{\frac{1}{2x}+\frac{3}{2(x+4)}-\frac{16}{3(4x-3)}\}$

Let $y=\frac{\sqrt{x}{(x+4)}^{\frac{3}{2}}}{{(4x-3)}^{\frac{4}{3}}}$

Taking $\log$ of both sides, we get

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

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

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

$\Rightarrow \frac{dy}{dx}=y\{\frac{1}{2x}+\frac{3}{2(x+4)}-\frac{16}{3(4x-3)}\}$

$=\frac{\sqrt{x}{(x+4)}^{\frac{3}{2}}}{{(4x-3)}^{\frac{4}{3}}}\{\frac{1}{2x}+\frac{3}{2(x+4)}-\frac{16}{3(4x-3)}\}$