How to Find the Derivative of f(x)^g(x): Function Raised to a Function
In differential calculus, you learn rules for finding the derivative of the sum, difference, product, quotient, and composition of two functions. But what about a function raised to the power of another function? A function of the form y = f(x)^g(x) may seem tricky to derive, however, there is a special technique called the logarithmic transformation that allows you to find the first, second, third derivatives, etc.
Given a function y of the form y = f(x)^g(x), you first take the natural logarithm of both sides. This gives you the equivalent expression
Ln(y) = Ln[f(x)^g(x)]
Ln(y) = g(x)*Ln(f(x))
The last expression is true because log(a^b) = b*log(a), one of the fundamental rules of logarithms. If we take the derivative of both sides of the last expression, we get
y'/y = g'(x)*Ln(f(x)) + g(x)*(1/f(x))*f'(x)
y'/y = g'(x)*Ln(f(x)) + g(x)*f'(x)/f(x)
y' = y[ g'(x)*Ln(f(x)) + g(x)*f'(x)/f(x) ]
y' = [ f(x)^g(x) ]*[ g'(x)*Ln(f(x)) + g(x)*f'(x)/f(x) ]
Here are some examples of this derivation trick in action.
Example 1: Find the Derivative of y = x^x
One of the simplest examples of such a function is y = x^x. Using the logarithmic derivative technique, we have
y = x^x
Ln(y) = x*Ln(x)
y'/y = Ln(x) + 1
y' = y[Ln(x) + 1]
y' = (x^x)[Ln(x) + 1]
If you want to find the critical points of this function on the positive real axis, you set y' equal to 0 and solve for x. This gives you
(x^x)[Ln(x) + 1] = 0
Ln(x) + 1 = 0
Ln(x) = -1
x = 1/e.
The point x = 1/e is the global minimum of y = x^x over the positive real numbers.
Example 2: Find the Derivative of y = cos(x)^sin(x)
For this function, it is useful to know that the derivative of cos(x) is -sin(x), and that the derivative of sin(x) is cos(x).
y = cos(x)^sin(x)
Ln(y) = sin(x)Ln(cos(x))
y'/y = cos(x)Ln(cos(x)) + sin(x)*(-sin(x)/cos(x))
y'/y = cos(x)Ln(cos(x)) - (sin(x)^2)/cos(x)
y' = [cos(x)^sin(x)]*[cos(x)Ln(cos(x)) - (sin(x)^2)/cos(x)]
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