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. $1$
• E. $-4$

Answer 1

Answer: (E)

$-4$

$y=(1-x)(1+x^2)(1+x^4)=(1-x+x^2-x^3)(1+x^4)$, multiply the first two terms

Now differentiate using product rule,

$\frac{dy}{dx}=(1-x+x^2-x^3)\left(4x^3\right)+(1+x^4)(-1+2x-3x^2)$

Substitute $x=1$ , to get the required value

$\frac{dy}{dx}{|}_{x=1}=(1-1+1-1)\left(4\right)+(1+1)(-1+2-3)=0-4=-4$