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How do you find the derivative of #arcsin(4x)#?

Answer 1

Paisley Johnson

# 4 / (sqrt{1 - 16x^2})#

#y = arcsin(4x)#

#sin y = 4x#

#cos y y' = 4#

#y' = 4/ (cos y)#

#= 4 / (sqrt{1 - sin^2 y})#

#= 4 / (sqrt{1 - (4x)^2})#

#= 4 / (sqrt{1 - 16x^2})#

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Answer 2

Isabella Ackerman

To find the derivative of (y = \arcsin(4x)), use the chain rule. The derivative of (\arcsin(u)) with respect to (u) is (\frac{1}{\sqrt{1-u^2}}), and then you multiply by the derivative of (u) with respect to (x), where (u = 4x).

So, start with:

[y = \arcsin(4x)]

Taking the derivative of both sides with respect to (x), we get:

[ \frac{dy}{dx} = \frac{1}{\sqrt{1-(4x)^2}} \cdot \frac{d}{dx}(4x) ]

[ \frac{dy}{dx} = \frac{1}{\sqrt{1-16x^2}} \cdot 4 ]

[ \frac{dy}{dx} = \frac{4}{\sqrt{1-16x^2}} ]

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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.