# How do you differentiate #y= 2 sin ^-1 (4x^4)#?

Use the chain rule:

When given:

Substitute into the chain rule:

Reverse the substitution for g:

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To differentiate ( y = 2\sin^{-1}(4x^4) ) with respect to (x), we'll use the chain rule and the derivative of the inverse sine function ((\sin^{-1}x) or ( \arcsin x)). The derivative of (\sin^{-1}x) with respect to (x) is (\frac{1}{\sqrt{1-x^2}}).

Given function: ( y = 2\sin^{-1}(4x^4) )

Differentiate both sides with respect to (x):

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

Apply the chain rule, (\frac{d}{dx}[\sin^{-1}(u)] = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}), where ( u = 4x^4 ).

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

Next, differentiate (u = 4x^4):

[ \frac{du}{dx} = 16x^3 ]

So, we have:

[ \frac{dy}{dx} = 2 \cdot \frac{1}{\sqrt{1-16x^8}} \cdot 16x^3 ]

Simplifying:

[ \frac{dy}{dx} = \frac{32x^3}{\sqrt{1-16x^8}} ]

This is the derivative of (y = 2\sin^{-1}(4x^4)) with respect to (x).

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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.

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.

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.

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