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First Principle of Differentiation

If y=sin-1x...

Question

If $y = {sin}^{- 1} (x^{2})$ then find $\frac{d y}{d x}$ using first principle.

A

\frac{2 x}{\sqrt{1 - x^{4}}}

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B

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

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C

\frac{x}{\sqrt{1 - x^{4}}}

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D

- \frac{1}{\sqrt{1 - x^{4}}}

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Solution

The correct option is B $\frac{2 x}{\sqrt{1 - x^{4}}}$
$y = {sin}^{- 1} (x^{2})$

$\frac{d y}{d x} = \frac{1}{\sqrt{1 - x^{4}}} \times 2 x$

$= \frac{2 x}{\sqrt{1 - x^{4}}}$

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Similar questions

y = {sin}^{- 1} (x^{2} \sqrt{1 - x} - \sqrt{x} \sqrt{1 - x^{4}}),

\frac{d y}{d x} .

\int x \sqrt{\frac{1 - x^{2}}{1 + x^{2}}} d x =

y = \frac{x^{4} + x^{2} + 2}{x^{2} - x + 1}

\frac{d y}{d x} = . . . . . . . . . . . . . . . . . . .

y = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + . . . . \infty

, then find the value of

\frac{d y}{d x}

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

\frac{d y}{d x}

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