Derivative with many exponents? Chain rule!?

How does this work?

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

soo confused

thanks!

2 Answers

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  • 8 years ago
    Favorite 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//

  • fizixx
    Lv 7
    8 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)

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