If y=e^{log _{e}left[1+x+x^{2}+ldotsright]},t | vedclass

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

Question

(B) Given,$y=e^{\log _{e} [1+x+x^{2}+\ldots]}$We know that the sum of an infinite geometric series is given by $1+x+x^{2}+\ldots = \frac{1}{1-x}$ for

Vedclass · Accepted Answer

$\frac{1}{(1-x)^{2}}$

Vedclass · Answer

(B) Given,$y=e^{\log _{e} [1+x+x^{2}+\ldots]}$We know that the sum of an infinite geometric series is given by $1+x+x^{2}+\ldots = \frac{1}{1-x}$ for $|x| Therefore,$y = e^{\log _{e} (1-x)^{-1}}$.Using the property $e^{\log _{e} f(x)} = f(x)$,we get $y = (1-x)^{-1}$.Now,differentiating with respect to $x$ using the chain rule:$\frac{d y}{d x} = \frac{d}{d x} (1-x)^{-1} = -1(1-x)^{-2} \cdot \frac{d}{d x}(1-x)$$\frac{d y}{d x} = -1(1-x)^{-2} \cdot (-1) = (1-x)^{-2} = \frac{1}{(1-x)^{2}}$.

If $y=e^{\log _{e}\left[1+x+x^{2}+\ldots\right]}$,then $\frac{d y}{d x}$ is equal to

Similar Questions

$\frac{d}{dx} \left( \log \left( \sqrt{x + \sqrt{x^2 + a^2}} \right) \right) = $

If $f(x) = \cot^{-1}\left(\frac{x^x - x^{-x}}{2}\right)$,then the value of $f'(1)$ is equal to

Differentiate the following with respect to $x$: $\log (\log x)$,where $x > 1$.

$\frac{d}{d x}(\log _{|x|} e) =$ . . . . . .

The differential coefficient of the function $\log_e \left( \sqrt{\frac{1 + \sin x}{1 - \sin x}} \right)$ with respect to $x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module