If fx - 1-2x-3x2- -x- - 1- then f-x-

The correct option is D 2(1+x)^3 Given f(x)= 1 2x+3x^2+.....∞ and |x| lt;1 The expansion of (1+x)^n=1+nx+n(n 1)/2x^2+.........+∞ So the ...

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Implicit Differentiation

If fx = 1-2...

Question

If $f (x) = 1 - 2 x + 3 x^{2} + \dots \infty . | x | < 1$ , then $f^{'} (x) =$

- \frac{1}{(1 - x)}

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

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

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- \frac{2}{(1 + x)^{3}}

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Solution

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f (x) = c o t^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}), g (x) = c o s^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})

f (x) = g (x)

x = \frac{\sqrt{25 - 10 \sqrt{5}}}{5}

s i n 18^{o} = \frac{\sqrt{5} - 1}{4}

\int \frac{x^{2} - x - 1}{\sqrt[3]{1 - 3 x}} d x = \frac{- (x^{2} - x - 1) {(1 - 3 x)}^{\frac{2}{3}}}{2} - \frac{(2 x - 1) {(1 - 3 x)}^{\frac{8}{3}}}{k} + C

k =

{cot}^{- 1} (\frac{1 - x^{2}}{2 x}) + {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = \frac{2 π}{3}, x > 0

x \neq 1

x =

f (x) = {cot}^{- 1} \frac{3 x - x^{3}}{1 - 3 x^{2}}

g (x) = {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}}

lim x \to \infty \frac{f (x) - f (a)}{g (x) - g (a)}

(0 < a < \frac{1}{2})

0 \leq x < 1

sin {{tan}^{- 1} \frac{1 - x^{2}}{2 x} + {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}}}

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