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

Answer 1

Charles Arocha

#dy/dx = (5x^4)/sqrt(1-x^10)#

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

#d/dx (arcsinx) = 1/sqrt(1-x^2)#

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

#d/dx (arcsinu) = 1/sqrt(1-u^2) (du)/dx#

So, we get:

For #y = arcsin(x^5)#,

#dy/dx = 1/sqrt(1-(x^5)^2) d/dx(x^5)#

# = (5x^4)/sqrt(1-x^10)#

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Answer 2

Adam Adkins

The derivative of ( y = \arcsin(x^5) ) is ( \frac{d}{dx}(\arcsin(x^5)) = \frac{1}{\sqrt{1 - (x^5)^2}} \cdot \frac{d}{dx}(x^5) ). Applying the chain rule, ( \frac{d}{dx}(x^5) = 5x^4 ). So, the derivative of ( y = \arcsin(x^5) ) is ( \frac{5x^4}{\sqrt{1 - x^{10}}} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.