If y=1+ (1/x) +(1/x2)+ (1/x3)+.......... to ∞ with |x|1 then (dy/dx)=

Question

If y=1+ (1/x) +(1/x2)+ (1/x3)+.......... to ∞ with |x|>1 then (dy/dx)= (A) (-y2/x2) (B) (y2/x2) (C) x2 y2 (D) (x2/y2)

Tardigrade Admin · Accepted Answer

We have y=1+(1/x)+(1/x2)+(1/x3)+ ldots y =(1/1-(1)x) (GP series) =(x/x-1) dots(i) On differentiating, we get (d y/d x)=((x-1)-x(+1)/(1-x)2)=-(1/(1-x)2) =-(y2/x2) (From (i))

Q. If $y = 1 + \frac{1}{x} + \frac{1}{x ^{2}} + \frac{1}{x ^{3}} + ..........$ to $\infty$ with $∣ x ∣ > 1$ then $\frac{d y}{d x}$ =

$\frac{- y ^{2}}{x ^{2}}$

$\frac{y ^{2}}{x ^{2}}$

$x^{2} y^{2}$

$\frac{x ^{2}}{y ^{2}}$

Q. If y=1+x1​+x21​+x31​+.......... to ∞ with ∣x∣>1 then dxdy​=

x2−y2​

x2y2​

x2y2

y2x2​

We have y=1+x1​+x21​+x31​+… y=1−x1​1​ (GP series) =x−1x​…(i) On differentiating, we get dxdy​=(1−x)2(x−1)−x(+1)​=−(1−x)21​ =−x2y2​ (From (i))

Q. If $y = 1 + \frac{1}{x} + \frac{1}{x ^{2}} + \frac{1}{x ^{3}} + ..........$ to $\infty$ with $∣ x ∣ > 1$ then $\frac{d y}{d x}$ =

$\frac{- y ^{2}}{x ^{2}}$

$\frac{y ^{2}}{x ^{2}}$

$x^{2} y^{2}$

$\frac{x ^{2}}{y ^{2}}$