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How do you differentiate #y= ln e^(6x+1)#?

Answer 1

Scarlett Allison

#dy/dx=6#.

#y=lne^(6x+1)#.

But, #lne^t=tlne=t*1=t#.

So, #y=6x+1#.

#:. dy/dx=6#.

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Answer 2

Axel Alvarez

To differentiate y = ln(e^(6x+1)), you can use the chain rule.

dy/dx = (d/dx) ln(e^(6x+1)) = (1/(e^(6x+1))) * (d/dx) (e^(6x+1)) = (1/(e^(6x+1))) * (6e^(6x+1))

So, dy/dx = 6/(e^(6x+1)).

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Answer 3

Paisley Johnson

To differentiate y = ln(e^(6x + 1)), we can use the chain rule. The derivative of ln(u) with respect to x is (1/u) * (du/dx), where u is a function of x. Here, u = e^(6x + 1).

First, differentiate the inner function, e^(6x + 1), with respect to x. The derivative of e^(u) with respect to u is e^(u), and then we multiply by the derivative of the exponent (6x + 1) with respect to x, which is 6.

So, du/dx = e^(6x + 1) * 6.

Then, using the chain rule, we have:

dy/dx = (1/u) * (du/dx) = (1/e^(6x + 1)) * (e^(6x + 1) * 6) = 6.

Therefore, the derivative of y = ln(e^(6x + 1)) with respect to x is simply 6.

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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.