# quick differentiation question

probably easy:

diff. 3^(2x) w.r.t.x

thanks
zoidberg
probably easy:

diff. 3^(2x) w.r.t.x

thanks

y = 3^2x

ln y = ln 3^2x

ln y = 2xln 3

Implicit differentiation for LHS: 1/y dy/dx = 2ln 3

dy/dx = y.2ln 3

dy/dx = 2ln 3.3^2x
zoidberg
probably easy:

diff. 3^(2x) w.r.t.x

thanks

let y=3^2x.
lny=2xln3
dy.dx=3^2x.2ln3.
Nylex
y = 3^2x

ln y = ln 3^2x

ln y = 2xln 3

Implicit differentiation for LHS: 1/y dy/dx = 2ln 3

dy/dx = y.2ln 3

dy/dx = 2ln 3.3^2x

So not that easy really...!
thanks
can u not use the chain rule for that?
I don't think you can because the index is a variable (contains x).
mik1a
I don't think you can because the index is a variable (contains x).

Yeah, that's right.
This is the standard y=a^x thingy thats one the syllabus isn't it?!? I seem to have a vague memory of being told that it was quite important....
Leekey
This is the standard y=a^x thingy thats one the syllabus isn't it?!?

Yep.
You dont actually need to do quite that,
3^(2x) = 9^x = e^(xln9)

dy/dx = ln9 * 9x = 2ln3 * 3^(2x)
JamesF
You dont actually need to do quite that,
3^(2x) = 9^x = e^(xln9)

dy/dx = ln9 * 9x = 2ln3 * 3^(2x)

I still prefer taking logs and then using the product rule.