What is the derivative of this function $y = \frac{1}{arcsin (2 x)}$ $y=1/arcsin(2x)$ ?

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What is the derivative of this function $y = \frac{1}{arcsin (2 x)}$ $y=1/arcsin(2x)$ ?

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Answer 1 · 2016-11-08T19:13:55

Start by differentiating $y = arcsin (2 x)$ $y = arcsin(2x)$ .

$y = arcsin (2 x) \to sin y = 2 x$ $y = arcsin(2x) -> siny = 2x$

Through implicit differentiation, we have:

$cos y (\frac{d y}{d x}) = 2$ $cosy(dy/dx) = 2$

$\frac{d y}{d x} = \frac{2}{cos y}$ $dy/dx = 2/(cosy)$

Apply the identity $cos y = \sqrt{1 - {sin}^{2} (y)}$ $cosy = sqrt(1 - sin^2(y))$

$\frac{d y}{d x} = \frac{2}{\sqrt{1 - {sin}^{2} y}} = \frac{2}{\sqrt{1 - 4 x^{2}}} \to since siny = 2x$ $dy/dx = 2/sqrt(1 - sin^2y) = 2/sqrt(1 - 4x^2) ->" since siny = 2x"$

So, now we can differentiate the entire expression using the quotient rule.

$\frac{d y}{d x} = \frac{0 \times arcsin (2 x) - \frac{2}{\sqrt{1 - 4 x^{2}}}}{{(arcsin (2 x))}^{2}}$ $dy/dx = (0 xx arcsin(2x) - 2/sqrt(1 - 4x^2))/(arcsin(2x))^2$

$\frac{d y}{d x} = - \frac{2}{(\sqrt{1 - 4 x^{2}}) {(arcsin (2 x))}^{2}}$ $dy/dx = -2/((sqrt(1 - 4x^2))(arcsin(2x))^2)$

Hopefully this helps!

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What is the derivative of this function $y = \frac{1}{arcsin (2 x)}$ $y=1/arcsin(2x)$ ?

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What is the derivative of this function y=1/arcsin(2x)y=1arcsin(2x)y=1/arcsin(2x)?

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What is the derivative of this function $y = \frac{1}{arcsin (2 x)}$ $y=1/arcsin(2x)$ ?