Question

If $y =(1+x)(1+x^2)(1+x^4),$ then $ \frac {dy}{dx}$ at $ x=1$ is

• A. 28
• B. 0
• C. 20
• D. 1

Answer 1

Answer: (A)

Given, $y=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)$
On differentiating both sides w.r.t. $x$, we get
$\frac{d y}{d x}=(1+x)\left(1+x^{2}\right) \frac{d}{d x}\left(1+x^{4}\right)$
$+(1+x)\left(1+x^{4}\right) \frac{d}{d x}\left(1+x^{2}\right)$
$+\left(1+x^{2}\right)\left(1+x^{4}\right) \frac{d}{d x}(1+x)$
$=(1+x)\left(1+x^{2}\right)\left(4 x^{3}\right)+(1+x)\left(1+x^{4}\right)(2 x)$
$+\left(1+x^{2}\right)\left(1+x^{4}\right)(1)$
$\Rightarrow \left(\frac{d y}{d x}\right)_{(x=1)}=(1+1)\left(1+1^{2}\right)\left(4 \times 1^{3}\right)$
$+(1+1)\left(1+1^{4}\right)(2 \times 1)+\left(1+1^{2}\right)\left(1+1^{4}\right)$
$=2 \times 2 \times 4+2 \times 2 \times 2+2 \times 2$
$=16+8+4=28$