Question

If $y=\frac{\sqrt[3]{31+3x}\sqrt[4]{41+4x}\sqrt[5]{51+5x}}{\sqrt[7]{71+7x}\sqrt[8]{81+8x}}$, then $y^{\prime }\left(0\right)$ is equal to

• A. $-1$
• B. $1$
• C. $2$
• D. Non existant

Answer 1

The correct option is B $1$

$y=\frac{\sqrt[3]{31+3x}\sqrt[4]{41+4x}\sqrt[5]{51+5x}}{\sqrt[7]{71+7x}\sqrt[8]{81+8x}}$
Take log both sides,
$\log y=\frac{1}{3}\log (1+3x)+\frac{1}{4}\log (1+4x)+\frac{1}{5}\log (1+5x)-\frac{1}{7}\log (1+7x)-\frac{1}{8}\log (1+8x)$
Differentiating both side w.r.t $x$
$\frac{1}{y}\times \frac{dy}{dx}=\frac{1}{(1+3x)}+\frac{1}{1+4x}+\frac{1}{1+5x}-\frac{1}{1+7x}-\frac{1}{1+8x}$
at $x=0,y=1$,

$\therefore \frac{dy}{dx}=1+1+1-1-1=1$