ln derivitives?

Question

3e^2x + Ln(3x)

find the derivitive

i've got no idea how to do this problem

Answer 1

You can take the derivative of each term separately, because of derivative properties.

So for the first term 3e^2x.... the derivative of e^x is always e^x, and the derivative of e^2x would be 2e^2x, because you have to remember the chain rule from the 2x. Therefore, the derivative of 3e^2x is 6e^2x.

And for the second term, the derivative of ln(x) is 1/x, which is 1/(the inside function). So for ln(3x), you have (1/3x)*(3) from the chain rule of the inside function, and it simplifies down to 1/x. You'll find that whenever you have to take the derivative of the ln(kx) [k is a constant], the derivative will always be 1/x.

So your final answer would be...
6e^(2x) + 1/x

Answer 2

I think it is this:

6e^2x + 1/x

I used the chain rule to get each term:

First term, based on d/dx(e^x) = e^x:

d/dx(3e^2x) = 3e^2x * d/dx(2x)
= 3e^2x * 2
= 6e^2x

Second term, based on d/dx(ln(x) = 1/x:

d/dx(ln3x) = 1/(3x) * d/dx(3x)
= 1/(3x) * 3
= 1/x

Now, something is wrong here because the integral of 1/x gives ln(x), not ln(3x).

Watch out, I think I'm wrong in the details. I'll have to think some more about this...
.
FOLLOWUP:

Oops, well, the other two guys backed me up. Looks like I was right after all. Good to see I still have it after all these years! Thanks guys!

Answer 3

Assuming you mean

f(x) = 3e^(2x) + ln(3x)

You solve the derivative by knowing your derivatives and how to apply the chain rule. The derivative of e^x is e^x, and the derivative of ln(x) is 1/x. In our case, it's not simply x we have that e is the power to, and it's not simply ln(x) that we're taking the derivative of, neither. We have to apply the chain rule.

f'(x) = 3e^(2x)[2] + [1/(3x)](3)
f'(x) = 6e^(2x) + 1/x

Answer 4

set the equation = to y and take the ln of both sides
y = 3e^2x + Ln(3x)
lny = ln(3e)^2x + ln(ln(3x))
you can then multiply by the exponent
lny = 2xln(3e) + ln(ln(3x))
now take the derivative
the derivative of a ln u is u’/u
y’/y = 2ln(3e) + 1/(xln(3x))
y' = (2ln(3e) + 1/(xln(3x)))y
y=3e^2x + Ln(3x) so
y' = (2ln(3e) + 1/(xln(3x)))(3e^2x + Ln(3x)
)
{the derivitive of ln lnu is u'/u/lnu}