The derivative of sin-1 2x-1 - x2 with respect to tan-1 2x-1 - x2 is

Let u = sin^ 1 ( 2x/1 + x^2 ) = 2 tan^ 1x ∴du/dx = 2/1 + x^2 and v = tan^ 1 ( 2x/1 + x^2 ) = 2 tan^ 1x ∴dv/dx = 2/1+x^2 ∴du/dv = ( du/dx )/ ( dv/dx ) = 1 ...

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Standard XIII

Mathematics

Derivative of One Function w.r.t Another

The derivativ...

Question

The derivative of $s i n^{- 1} (\frac{2 x}{1 + x^{2}})$ with respect to $t a n^{- 1} (\frac{2 x}{1 - x^{2}})$ is

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

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

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Solution

The correct option is B $1$
$L e t u = s i n^{- 1} (\frac{2 x}{1 + x^{2}}) = 2 t a n^{- 1} x$
$∴ \frac{d u}{d x} = \frac{2}{1 + x^{2}}$
$a n d v = t a n^{- 1} (\frac{2 x}{1 + x^{2}}) = 2 t a n^{- 1} x$
$∴ \frac{d v}{d x} = \frac{2}{1 + x^{2}}$
$∴ \frac{d u}{d v} = \frac{(\frac{d u}{d x})}{(\frac{d v}{d x})} = 1$

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Similar questions

Q. The derivative of

s i n^{- 1} (\frac{2 x}{1 + x^{2}})

with respect to

t a n^{- 1} (\frac{2 x}{1 - x^{2}})

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{tan}^{- 1} \frac{2 x}{1 - x^{2}}

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s i n^{- 1} (\frac{2 x}{1 + x^{2}})

with respect to

t a n^{- 1} (\frac{2 x}{1 + x^{2}})

Q. The derivative of

s i n^{- 1} (\frac{2 x}{1 + x^{2}})

with respect to

t a n^{- 1} (\frac{2 x}{1 - x^{2}})

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The derivative of sin−1(2x1+x2) with respect to tan−1(2x1−x2) is

The correct option is B 1Let u=sin−1(2x1+x2)=2 tan−1x ∴dudx=21+x2 and v=tan−1(2x1+x2)=2 tan−1x ∴dvdx=21+x2 ∴dudv=(dudx)(dvdx)=1

The derivative of $s i n^{- 1} (\frac{2 x}{1 + x^{2}})$ with respect to $t a n^{- 1} (\frac{2 x}{1 - x^{2}})$ is