Question

If $\frac{d}{d x}\left(\frac{1 + x^{4} + x^{8}}{1 + x^{2} + x^{4}}\right)=ax^{3}+bx$ , then find the value of $a+b.$

Answer 1

Let $y=\frac{1 + x^{4} + x^{3}}{1 + x^{2} + x^{4}}$
$=\frac{\left(1 + x^{4}\right)^{2} - x^{4}}{1 + x^{2} + x^{4}}$
$=\frac{\left(1 + x^{4}\right)^{2} - \left(x^{2}\right)^{2}}{1 + x^{2} + x^{4}}$
$=\frac{\left(1 + x^{4} + x^{2}\right) \left(1 + x^{4} - x^{2}\right)}{1 + x^{2} + x^{4}}$
$\Rightarrow y=1+x^{4}-x^{2}$
$\Rightarrow \frac{d y}{d x}=4x^{3}-2x$
$\Rightarrow a=4,b=-2$
$\Rightarrow a+b=4-2=2$