If y=(1+x)(1+x^2)(1+x^4) dots (1+x^{2^n}),the | vedclass

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

Question

(C) Given $y = (1+x)(1+x^2)(1+x^4) \dots (1+x^{2^n})$.Taking the natural logarithm on both sides:$\ln y = \ln(1+x) + \ln(1+x^2) + \ln(1+x^4) + \dots +

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Given $y = (1+x)(1+x^2)(1+x^4) \dots (1+x^{2^n})$.Taking the natural logarithm on both sides:$\ln y = \ln(1+x) + \ln(1+x^2) + \ln(1+x^4) + \dots + \ln(1+x^{2^n})$.Differentiating both sides with respect to $x$:$\frac{1}{y} \frac{dy}{dx} = \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \dots + \frac{2^n x^{2^n-1}}{1+x^{2^n}}$.Thus,$\frac{dy}{dx} = y \left[ \frac{1}{1+x} + \frac{2x}{1+x^2} + \dots + \frac{2^n x^{2^n-1}}{1+x^{2^n}} \right]$.At $x=0$,$y = (1+0)(1+0) \dots (1+0) = 1$.Substituting $x=0$ into the derivative expression:$\left(\frac{dy}{dx}\right)_{x=0} = 1 \cdot \left[ \frac{1}{1+0} + 0 + 0 + \dots + 0 \right] = 1 \cdot 1 = 1$.

If $y=(1+x)(1+x^2)(1+x^4) \dots (1+x^{2^n})$,then the value of $\left(\frac{dy}{dx}\right)_{x=0}$ is

Similar Questions

State the electronic configurations of group-$16$ elements.

The element with atomic number $12$ belongs to ......... group and ......... period.

If $\alpha, \beta$ and $\gamma$ are the lengths of the altitudes of a $\triangle ABC$ with area $\Delta$,then $\frac{\Delta^2}{R^2}\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}\right)$ is equal to

Suppose $E$ and $F$ are two events of a random experiment. If the probability of occurrence of $E$ is $1/5$ and the probability of occurrence of $F$ given $E$ is $1/10$,then the probability of non-occurrence of at least one of the events $E$ and $F$ is

In the equation $log\, S = log\, C + Z\, log\, A$,what does $S$ represent?

Vedclass Test Series

Exam Paper Generator

Online Exam Module