Question

If y=1+(1/x)+(1/x^2)+(1/x^3)+.... infinity with|x|>1, then dy/dx is
A. x^2/y^2
B. x^2.y^2
C. y^2/x^2
D. -y^2/x^2

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

here we have given a series ,to find the derivative of given function, first we have to find the sum of given series

$\sf y=1+(\frac{1}{x})+(\frac{1}{x^2})+(\frac{1}{x^3})+.......∞$

In a Geometric series $\sf (\frac{tn+1}{tn})$is a constant, which is called the common difference n€ N

let's check the common difference of the given series,

1+(1/x)+(1/x²)+(1/x³)+....∞

here, tn=1 and tn+1=(1/x)

common difference =(tn+1)/tn =(1/x)/1 = (1/x)............(1)

for, the second one ,

tn= (1/x) ,tn+1=(1/x²)

common difference =(tn+1)/tn =(1/x²)/(1/x)=x/x²=(1/x).....(2)

from (1) and (2) given series is geometric progression.

where, a = first term and r = common difference = (1/x)

here first, we have to find the sum of series tends to infinity ,

generally sn in Gp ,

$\sf Sn=[\frac{a(1-r^n)}{1-r}]=(\frac{a}{1-r})-(\frac{a}{1-r})r^n$

here, sum is tends to infinity so if, |r|<1 ,n is large , r^n is taken as zero so the formula will be $\sf(\frac{a}{1-r})$

Now, let's find out the sum of given series [ a = 1 , r = (1/x)]

$\implies\sf (\frac{a}{1-r})=(\frac{1}{1-(\frac{1}{x})})$

$\implies \sf (\frac{x}{x-1})$

so the function becomes,

$y = \sf (\frac{x}{x-1})$

Differentiate wrt x

$\sf \frac{dy}{dx}=(\frac{d}{dx}(\frac{x}{x-1}))$

here use division rule for differentiation ,

if y = u/v is the function then their derivative is dy/dx=[(du/dx)v-(dv/dx)u]/v² therefore,

$\implies \sf \frac{1(x-1)-(1-0)x}{(x-1)^2}$

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

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

Now, we have ,

$y = \sf (\frac{x}{x-1})$

$\implies \sf \frac{dy}{dx}=\frac{-1}{(x-1)^2}$

in the option we have x² term, so multiply denominator and numerator by x² on RHS ,

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

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

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

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

$\implies \sf -(\frac{(y)^2}{x^2})$

Hence option D is correct