y=sin^ 1(2x1+x^2) Now, here we are using chain rule. dydx=ddxsin^ 1(2x1+x^2) #160; #160; #160; #160; =cos(2x1+x^2)ddx(2x1+x^2) #160; ...

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Chain Rule of Differentiation

Solve y=sin...

Question

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

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Solution

$y = {sin}^{- 1} (\frac{2 x}{1 + x^{2}})$
Now, here we are using chain rule.
$\frac{d y}{d x} = \frac{d}{d x} {sin}^{- 1} (\frac{2 x}{1 + x^{2}})$

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

$= cos (\frac{2 x}{1 + x^{2}}) ⎛ ⎜ ⎜ ⎜ ⎝ \frac{(1 + x^{2}) \frac{d}{d x} (2 x) - 2 x \frac{d}{d x} (1 + x^{2})}{(1 + x^{2})^{2}} ⎞ ⎟ ⎟ ⎟ ⎠$

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

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

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

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