Question

If |x|<1 and y=1+x+x^2+...to infinity, then find the value of dy/dx .

Answer 1

$Answer \: \\ \\ GIVEN \: \: QUESTION \: \: Is \: \: \\ \\ y = 1 + x + x {}^{2} + x {}^{3} + x {}^{4} + ... + \infty \\ \\ The \: \: above \: series \: is \: in \: gp \: becoz \: \\ common \: ratio \: is \: same \: \\ \\ its \: sum \: is \: given \: by \: \: \: \: \frac{a}{(1 - r)} \: \: \: \: r < 1 \\ \\ here \: \: a \: = 1 \: \: \: and \: \: \: r = \frac{x {}^{2} }{x} = x \\ so \\ y = \frac{1}{(1 - x)} \\ \\ now \: \: differentiate \: both \: sides \: with \: \\ respect \: to \: x \: we \: have \\ \\ \frac{dy}{dx} = \frac{ - 1}{(1 - x) {}^{2} } \times - 1 \\ \\ \frac{dy}{dx} = \frac{1}{(1 - x) {}^{2} }$

Answer 2

$Answer \: \\ \\ GIVEN \: \: QUESTION \: \: Is \: \: \\ \\ y = 1 + x + x {}^{2} + x {}^{3} + x {}^{4} + ... + \infty \\ \\ The \: \: above \: series \: is \: in \: gp \: becoz \: \\ common \: ratio \: is \: same \: \\ \\ its \: sum \: is \: given \: by \: \: \: \: \frac{a}{(1 - r)} \: \: \: \: r < 1 \\ \\ here \: \: a \: = 1 \: \: \: and \: \: \: r = \frac{x {}^{2} }{x} = x \\ so \\ y = \frac{1}{(1 - x)} \\ \\ now \: \: differentiate \: both \: sides \: with \: \\ respect \: to \: x \: we \: have \\ \\ \frac{dy}{dx} = \frac{ - 1}{(1 - x) {}^{2} } \times - 1 \\ \\ \frac{dy}{dx} = \frac{1}{(1 - x) {}^{2} }$