If $\displaystyle y= \frac {\sqrt[3]{1+3x}\sqrt[4]{1+4x}\sqrt[5]{1+5x}}{\sqrt[7]{1+7x}\sqrt[8]{1+8x}}$, then $y'(0)$ is equal to

$\displaystyle y= \frac {\sqrt[3]{1+3x}\sqrt[4]{1+4x}\sqrt[5]{1+5x}}{\sqrt[7]{1+7x}\sqrt[8]{1+8x}}$

Take log both sides,

$\displaystyle \log y = \frac{1}{3}\log(1+3x)+\frac{1}{4}\log(1+4x)+\frac{1}{5}\log(1+5x)-\frac{1}{7}\log(1+7x)-\frac{1}{8}\log(1+8x)$

Differentiating both side w.r.t $x$

$\displaystyle \frac {1}{y}\times \frac {dy}{dx}=\frac {1}{(1+3x)}+\frac {1}{1+4x}+\frac {1}{1+5x}-\frac {1}{1+7x}-\frac {1}{1+8x}$

at $x=0, y=1$,

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