Chain Rule Differentiation Problem

What is the derivative of 3^(7^(x^2)) ?
I tried logarithmic differentiation as well, but wasn't getting the right result, so I'm trying to use chain rule.

So far I have
u(x) = 3^7
g(x) = u^x
h(x) = (u^x)^2

I really want to learn how to do this question.... thanks guys, I'll def pick the best answer

You need to use the Chain Rule twice.

Let y = 3^[7^(x^2)] and u = 7^(x^2). Hence y = 3^u

dy/du = 3^(u)ln(3) = 3^[7^(x^2)] ln(3)

To find du/dx use the Chain Rule for the first time thus:

Let Y = 7^(x^2) and let U = x^2. Then Y = 7^U. Hence:

dY/dU = 7U ln(7) = 7x^(2)ln(7) and DU/dx = 2x

Chain Rule: dY/dx = dY/dU/* dU/dx = 7x^(2)ln(7)*2x

Hence du/dx = 2x*7x^(2)ln(7)

Using the Chain Rule again: dy/dx = dy/du * du/dx.

dy/dx = 3^[7^(x^2)]ln(3) * 2x*7x^(2)ln(7)

dy/dx = 2x*3^[7^x^(2)]* 7^x^(2)*ln(3)*ln(7)

Using the chain rule, d/dx (3^(7x^2)=3^u du/dx ln(3),

where u=7^(x^2) and d 3^u/du= 3^u ln(3):

3^(7x^2) ln(3) (d/dx (7^(x^2))

Using the chain rule, d/dx (7^(x^2))= 7^u du/dx ln(7),

where u=x^2 and d 7^u/du= 7^u ln(7):

3^(7^(x^2)) ln(3) (7x^2 ln(7) (d/dx (x^2)))

The derivative of x^2 is 2x:

Answer: 3^(7^(x^2)) 7^(x^2) (2x) ln(3) ln(7).

y = 3^(7^(x^2))
dy/dx = 3^[7^(x^2)]ln(3) d/dx 7^(x^2)ln(7) d/dx 2x
= 2x* 3^[7^(x^2)] * 7^(x^2)ln(3)ln(7) answer//