For |x|<1,let y=1+x+x2+… to ∞, then dydxis equal to

For |x|

A
$\frac{x}{y}$
B
$\frac{x^{2}}{y^{2}}$
C
$\frac{x}{y^{2}}$
D
${xy}^{2} + y$

Solution:

$\begin{array}{l} ∵ y = 1 + x + x^{2} + \dots \infty \\ ∴ y = \frac{1}{1 - x} = (1 - x)^{- 1} \end{array}$

On differentiating w.r.t. x, we get

$\begin{array}{l} \frac{dy}{dx} = - \frac{1}{(1 - x)^{2}} (- 1) = \frac{1}{(1 - x)^{2}} \\ ∴ \frac{dy}{dx} - y = \frac{1}{(1 - x)^{2}} - \frac{1}{(1 - x)} = \frac{1 - 1 + x}{(1 - x)^{2}} = \frac{x}{(1 - x)^{2}} \\ \Rightarrow \frac{dy}{dx} - y = {xy}^{2} \Rightarrow \frac{dy}{dx} = {xy}^{2} + y \end{array}$

For $| x | < 1, let y = 1 + x + x^{2} + \dots to \infty,$ then $\frac{dy}{dx}$ is equal to

Solution:

Related content