How do you differentiate $y = {({sin}^{- 1} (5 x^{2}))}^{3}$ $y=(sin^-1(5x^2))^3$ ?

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How do you differentiate $y = {({sin}^{- 1} (5 x^{2}))}^{3}$ $y=(sin^-1(5x^2))^3$ ?

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Answer 1 · 2016-11-06T19:02:33

$\frac{d y}{d x} = \frac{30 x {(arcsin (5 x^{2}))}^{2}}{\sqrt{1 - 5 x^{2}}}$ $\fracdydx=\frac{30x(arcsin(5x^2))^2}{\sqrt{1-5x^2}}$

Explanation:

The derivative of $arcsin (u)$ $\arcsin(u)$ (or ${sin}^{- 1} (u)$ $sin^-1(u)$ ) is $\frac{1}{\sqrt{1 - u^{2}}} \frac{d u}{d x}$ $\frac{1}{\sqrt{1-u^2}}\frac{du}dx$

So by using the chain rule,
$\frac{d y}{d x} = 3 {(arcsin (5 x^{2}))}^{2} \cdot \frac{1}{\sqrt{1 - 5 x^{2}}} \cdot 10 x$ $\fracdydx=3(arcsin(5x^2))^2\cdot\frac1{\sqrt{1-5x^2}}\cdot10x$

Which can be simplified to:
$\frac{d y}{d x} = \frac{30 x {(arcsin (5 x^{2}))}^{2}}{\sqrt{1 - 5 x^{2}}}$ $\fracdydx=\frac{30x(arcsin(5x^2))^2}{\sqrt{1-5x^2}}$

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How do you differentiate $y = {({sin}^{- 1} (5 x^{2}))}^{3}$ $y=(sin^-1(5x^2))^3$ ?

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How do you differentiate $y = {({sin}^{- 1} (5 x^{2}))}^{3}$ $y=(sin^-1(5x^2))^3$ ?