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Question

Find $\frac{d y}{d x}$ in the following: $y = sin^{- 1} (\frac{2 x}{1 + x ^{2}})$

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Find

\frac{d y}{d x}

in the following:

y = sin^{- 1} (\frac{2 x}{1 + x ^{2}})

11 mins ago

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Text solutionVerified

Solution:

y = sin^{- 1} (\frac{2 x}{1 + x ^{2}})

\Rightarrow sin y = \frac{2 x}{1 + x ^{2}}

\Rightarrow cos y = 1 - sin^{2} y = 1 - (\frac{2 x}{1 + x ^{2}})^{2} = \frac{1 - x ^{2}}{1 + x ^{2}}

Differentiating it w.r.t.

x

,

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

\Rightarrow cos y \frac{d y}{d x} = \frac{2 ( 1 - x ^{2} )}{( 1 + x ^{2} ) ^{2}}

\Rightarrow \frac{d y}{d x} = \frac{1}{cos y} (\frac{2 ( 1 - x ^{2} )}{( 1 + x ^{2} ) ^{2}})

\Rightarrow \frac{d y}{d x} = \frac{1 + x ^{2}}{1 - x ^{2}} * (\frac{2 ( 1 - x ^{2} )}{( 1 + x ^{2} ) ^{2}})

\Rightarrow \frac{d y}{d x} = \frac{2}{1 + x ^{2}}

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Question Text	Find $\frac{d y}{d x}$ in the following: $y = sin^{- 1} (\frac{2 x}{1 + x ^{2}})$
Answer Type	Text solution:1
Upvotes	150