Question

If $y=e^{{\log }_e\{1+x+x^2+x^3+\cdots \}}$, then $\frac{dy}{dx}=$

• A. $-\frac{1}{(1-x{)}^2}$
• B. $\frac{1}{(1+x{)}^2}$
• C. $\frac{1}{(1-x{)}^2}$
• D. $-\frac{1}{(1+x{)}^2}$

Answer 1

Answer: (C)

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

The given equation is:

$y=e^{{\log }_e\left(1+x+x^2+x^3+.....\right)}$

Using the property $y=e^{{\log }_{e}x}=x$ we get,

$y=1+x+x^2+x^3+.....$

The above equation is a G.P. with common ratio x. Hence the summation is,

$y=\frac{1}{1-x}$

Differentiating once w.r.t. to x we get,

$\frac{dy}{dx}=\frac{1}{(1-x{)}^2}$ ....Answer