What is the derivative of $y = arcsin (x^{5})$ $y = arcsin(x^5)$ ?

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Answer 1 · 2015-08-29T13:37:26

$\frac{d y}{d x} = \frac{5 x^{4}}{\sqrt{1 - x^{10}}}$ $dy/dx = (5x^4)/sqrt(1-x^10)$

Use the derivative of $arcsin$ $arcsin$ and the chain rule.

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

When we don't have $x$ $x$ as the argument of the function, then we need the chain rule:

$\frac{d}{d x} (arcsin u) = \frac{1}{\sqrt{1 - u^{2}}} \frac{d u}{d x}$ $d/dx (arcsinu) = 1/sqrt(1-u^2) (du)/dx$

So, we get:

For $y = arcsin (x^{5})$ $y = arcsin(x^5)$ ,

$\frac{d y}{d x} = \frac{1}{\sqrt{1 - {(x^{5})}^{2}}} \frac{d}{d x} (x^{5})$ $dy/dx = 1/sqrt(1-(x^5)^2) d/dx(x^5)$

$= \frac{5 x^{4}}{\sqrt{1 - x^{10}}}$ $= (5x^4)/sqrt(1-x^10)$

What is the derivative of y = arcsin(x^5)y=arcsin(x5)y = arcsin(x^5)?