How do you find the derivative of $arcsin (2 x^{2})$ $arcsin(2x^2)$ ?

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How do you find the derivative of $arcsin (2 x^{2})$ $arcsin(2x^2)$ ?

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Answer 1 · 2018-03-09T16:34:46

$\frac{4 x}{\sqrt{1 - 4 x^{4}}}$ $(4x)/sqrt(1-4x^4$

Explanation:

Here,
$y = {sin}^{- 1} (2 x^{2})$ $y=sin^(-1)(2x^2)$ , take , $u = 2 x^{2}$ $u=2x^2$
$y = {sin}^{- 1} u$ $y=sin^(-1)u$
$\frac{d y}{d u} = \frac{1}{\sqrt{1 - u^{2}}} and \frac{d u}{d x} = 4 x$ $(dy)/(du)=1/sqrt(1-u^2)and(du)/(dx)=4x$
$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x} = \frac{1}{\sqrt{1 - u^{2}}} \cdot 4 x$ $color(red)((dy)/(dx)=(dy)/(du)*(du)/(dx))=1/sqrt(1-u^2)*4x$
$\Rightarrow \frac{d y}{d x} = \frac{1}{\sqrt{1 - {(2 x^{2})}^{2}}} \cdot 4 x = \frac{4 x}{\sqrt{1 - 4 x^{4}}}$ $=>(dy)/(dx)=1/(sqrt(1-(2x^2)^2))*4x=(4x)/sqrt(1-4x^4$

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How do you find the derivative of $arcsin (2 x^{2})$ $arcsin(2x^2)$ ?

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How do you find the derivative of $arcsin (2 x^{2})$ $arcsin(2x^2)$ ?