If y -tan -11-1-x-x2-tan -11-x2-3 x-3-tan -11-x2-5 x-7-n terms- then y-0 isA- n2-1 n n2B- 2 n2-1-n2C- -n2-1-n2D- - - 22

The correct option is C tan^ 1(1/1+x+x^2 )=tan^ 1(x+1 x/1+x+(x+1) )=tan^ 1(x 1) tan^ 1x similarly other terms ...

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Byju's Answer

Standard XII

Mathematics

Differentiation of Inverse Trigonometric Functions

If y =tan -11...

Question

If $y = t a n^{- 1} (\frac{1}{1 + x + x^{2}}) + t a n^{- 1} (\frac{1}{x^{2} + 3 x + 3}) + t a n^{- 1} (\frac{1}{x^{2} + 5 x + 7}) + \dots \dots \dots$ n terms, then y'(0) is

π/2

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Solution

The correct option is D

$t a n^{- 1} (\frac{1}{1 + x + x^{2}}) = t a n^{- 1} (\frac{x + 1 - x}{1 + x + (x + 1)}) = t a n^{- 1} (x - 1) - t a n^{- 1} x$ similarly other terms

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Similar questions

Q. If

y = t a n^{- 1} \frac{1}{1 + x + x^{2}} + t a n^{- 1} \frac{1}{x^{2} + 3 x + 3} + t a n^{- 1} \frac{1}{x^{2} + 5 x + 7} + \dots +

upto n terms, then

y^{'} (0)

is equal to

Q. If

y = t a n^{- 1} (\frac{1}{1 + x + x^{2}}) + t a n^{- 1} (\frac{1}{x^{2} + 3 x + 3}) + t a n^{- 1} (\frac{1}{x^{2} + 5 x + 7}) + . . . . +

up to n terms. Then y’(0) is equal to

If $y = {tan}^{- 1} (\frac{1}{1 + x + x^{2}}) + {tan}^{- 1} (\frac{1}{x^{2} + 3 x + 3}) + {tan}^{- 1} (\frac{1}{7 + 5 x + x^{2}}) + \dots n$ terms, then ${(\frac{d y}{d x})}_{x = 0} =$

I f y = t a n^{- 1} (\frac{1}{1 + x + x^{2}}) + t a n^{- 1} (\frac{1}{x^{2} + 3 x + 3}) + t a n^{- 1} (\frac{1}{x^{2} + 5 x + 7})

+ ...... + up to n terms. Then y' (0) is equal to

Q. If

y = {tan}^{- 1} (\frac{1}{1 + x + x^{2}}) + {tan}^{- 1} (\frac{1}{x^{2} + 3 x + 3}) + {tan}^{- 1} (\frac{1}{x^{2} + 5 x + 7}) + . . . . n

terms then

y^{'} (0)

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If y=tan−1(11+x+x2)+tan−1(1x2+3x+3)+tan−1(1x2+5x+7)+⋯⋯⋯ n terms, then y'(0) is

The correct option is D tan−1(11+x+x2)=tan−1(x+1−x1+x+(x+1))=tan−1(x−1)−tan−1x similarly other terms

If $y = t a n^{- 1} (\frac{1}{1 + x + x^{2}}) + t a n^{- 1} (\frac{1}{x^{2} + 3 x + 3}) + t a n^{- 1} (\frac{1}{x^{2} + 5 x + 7}) + \dots \dots \dots$ n terms, then y'(0) is

The correct option is D

$t a n^{- 1} (\frac{1}{1 + x + x^{2}}) = t a n^{- 1} (\frac{x + 1 - x}{1 + x + (x + 1)}) = t a n^{- 1} (x - 1) - t a n^{- 1} x$ similarly other terms