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Calculus/Calculus Finding derivative

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Question
If y = x^(x^3) for x > 0, then find dy/dx

Answer
This is almost like a product rule, except that it is an exponential problem.  Read slow and understand.

The derivative of f(x)^g(x) is like looking at f(x)^n and m^g(x).

Here, f(x) = x and g(x) = x^3.
In the first one, n is really g(x).
In the second one, m is really f(x).

The two parts of the derivative would be
[nf(x)^(n-1)]f'(x) + [m^g(x)]ln(m)g'(x).

If we put back in what n amd m are, we get
[g(x)f(x)^(g(x)-1)]f'(x) + [f(x)^g(x)]g'(x)ln(f(x)).

In the write order it would be
g(x)f'(x)f(x)^(g(x)-1) + ln(f(x))g'(x)f(x)^g(x).

Basically, anyway you have two functions, treat them each individually when taking the derivative.

It may take a few minutes to read the problem and understand.
I could always explain it furtheer.

Calculus

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