Question

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

• A. $20$
• B. $28$
• C. $1$
• D. $0$

Answer 1

The correct option is B

$28$

Explanation for the correct option.

Step 1. Differentiate $y$ with respect to $x.$

Given an equation $y=\left(1+x\right)\left(1+x^2\right)\left(1+x^4\right)$. Now using the product rule of differentiation, differentiate both sides with respect to $x$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d}{dx}\left(\left(1+x\right)\left(1+x^2\right)\left(1+x^4\right)\right)\\ & =& 1·\left(1+x^2\right)\left(1+x^4\right)+\left(1+x\right)·2x·\left(1+x^4\right)+\left(1+x\right)\left(1+x^2\right)·4x^3\\ & =& \left(1+x^2\right)\left(1+x^4\right)+2x\left(1+x\right)\left(1+x^4\right)+4x^3\left(1+x\right)\left(1+x^2\right)\end{array}$

Step 2. Find the value of $\frac{dy}{dx}$ at $x=1$

In the expression $\begin{array}{rcl}& & \left(1+x^2\right)\left(1+x^4\right)+2x\left(1+x\right)\left(1+x^4\right)+4x^3\left(1+x\right)\left(1+x^2\right)\end{array}$ substitute $x=1$.

$\begin{array}{rcl}{\overline{)\frac{dy}{dx}}}_{x=1}& =& \left(1+1^2\right)\left(1+1^4\right)+2\times 1\left(1+1\right)\left(1+1^4\right)+4\times 1^3\left(1+1\right)\left(1+1^2\right)\\ & =& 2\times 2+2\times 2\times 2+4\times 2\times 2\\ & =& 4+8+16\\ & =& 28\end{array}$

Hence, the correct option is B.