If f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8),then f^{pr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)$.Taking the natural logarithm on both sides:$\ln(f(x)) = \ln(1+x) + \ln(1+x^2) + \ln(1+x^4) + \ln(1+x^8)$.D

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(D) Given $f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)$.Taking the natural logarithm on both sides:$\ln(f(x)) = \ln(1+x) + \ln(1+x^2) + \ln(1+x^4) + \ln(1+x^8)$.Differentiating both sides with respect to $x$:$\frac{f^{\prime}(x)}{f(x)} = \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8}$.At $x=1$,$f(1) = (1+1)(1+1)(1+1)(1+1) = 2 	imes 2 	imes 2 	imes 2 = 16$.Substituting $x=1$ into the derivative expression:$\frac{f^{\prime}(1)}{f(1)} = \frac{1}{1+1} + \frac{2(1)}{1+1^2} + \frac{4(1)^3}{1+1^4} + \frac{8(1)^7}{1+1^8} = \frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2} = \frac{1+2+4+8}{2} = \frac{15}{2}$.Therefore,$f^{\prime}(1) = f(1) 	imes \frac{15}{2} = 16 	imes \frac{15}{2} = 8 	imes 15 = 120$.

If $f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)$,then $f^{\prime}(1)=$

Similar Questions

If $y = x^{\ln x}$,then $dy/dx$ equals :-

If $h(x) = x^{x^x}$,then at $x = 1$,$\frac{h'(x)}{h(x)}$ is equal to

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

If $x^y=y^{\sin x}(\tan x)^{\cos x}$, then $\left(\log x-\frac{\sin x}{y}\right) \frac{d y}{d x}=$

If $y = \frac{x^2}{(x - 1)(x - 2)(x - 3)} + \frac{2x}{(x - 2)(x - 3)} + \frac{3}{x - 3} + 1$,then $\frac{xy'}{y}$ is equal to (where $y' = \frac{dy}{dx}$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module