# How do you differentiate #y=ln ln(2x^4)#?

Through the chain rule, we see that:

So, here, we see that:

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To differentiate ( y = \ln(\ln(2x^4)) ), we use the chain rule:

- First, differentiate the outer function, ln(u), where u = ln(2x^4), which gives 1/u.
- Then, multiply by the derivative of the inner function, u = ln(2x^4), which involves applying the chain rule again.

Derivative of the inner function u: [ u = \ln(2x^4) ] [ u' = \frac{1}{u} \cdot \left(\frac{1}{2x^4}\right) ]

Now, putting it all together: [ y' = \frac{1}{\ln(2x^4)} \cdot \frac{1}{2x^4} ]

So, the derivative of ( y = \ln(\ln(2x^4)) ) is: [ y' = \frac{1}{\ln(2x^4)} \cdot \frac{1}{2x^4} ]

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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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