If y =e loge[1+x+x2+.........], then (dy/dx)=

Question

If y =e loge[1+x+x2+.........], then (dy/dx)= (A)  (1/(1+x)2)  (B)  (1/(1-x)2)  (C)  (-1/(1+x)2)  (D)  (-1/(1-x)2)

Tardigrade Admin · Accepted Answer

Given, y = elog [1+x+x2+ ldots] ∵ (1-x)-1 = 1+x+x2+x3+ ldots∞ ∴ y = elog(1-x)-1 ⇒ y=(1-x)-1 On differentiating w.r.t. 'x', we get (dy/dx) = (-1)(1-x)-2 = (-1/(1-x)2)

Q. If $y = e^{l o g_{e} [1 + x + x^{2} + .........]}$ , then $\frac{d y}{d x}$ =

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

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

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

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

Given, $y = e^{l o g [1 + x + x^{2} + \dots]}$
$∵ (1 - x)^{- 1} = 1 + x + x^{2} + x^{3} + \dots \infty$
$∴ y = e^{l o g (1 - x)^{- 1}}$
$\Rightarrow y = (1 - x)^{- 1}$
On differentiating w.r.t. 'x', we get
$\frac{d y}{d x} = (- 1) (1 - x)^{- 2}$
$= \frac{- 1}{( 1 - x ) ^{2}}$

Q. If y=eloge​[1+x+x2+.........], then dxdy​=

(1+x)21​

(1−x)21​

(1+x)2−1​

(1−x)2−1​