# Derivative with many exponents? Chain rule!?

How does this work?

5^(7^(t^(3)))

soo confused

thanks!

### 2 Answers

- Engr. RonaldLv 78 years agoFavorite Answer
5^(7^(t^(3)))

=5^(7^(t^(3)))ln(5) d/dx 7^(t^3)ln(7) d/dx 3t^2

=3t^2 * 5^(7^(t^3))) * 7^(t^3) ln(5)ln(7) answer//

- fizixxLv 78 years ago
Think of it like this:

7^t^(3) = 7^u

And 5^7^u = 5^w

So ---- w = 7^u, and u = t^3

So....you have a function --- g and you need a derivative of t of that function, so you have to take:

(dg/dw)(dw/du)(du/dt)

derivative is: dg/dw ----- remember Ln(g) = wLn(5), so, taking the derivative is (1/f)(dg/dw) = Ln(5)

=> dg/dw = gLn(5) = (5^w)Ln(5)

Now dw/du ----------- w = 7^u

=> (7^u)Ln(7) by the same technique we just did above.

Then du/dt ------------ u = t^3

=> (3t^2)

Then what you end up with is: (dg/dw)(dw/du)(du/dt)

=> (5^w)Ln(5) (7^u)Ln(7) (3t^2)

simplify

=> (5^w)(7^u)(3t^2)Ln(5)Ln(7)

Put back in for 'w' and 'u'

=> (5^7^t^3)(7^t^3)(3t^2)Ln(5)Ln(7)