# Derivative of Power of Function/Proof 2

## Theorem

Let $\map u x$ be a differentiable real function of $x$.

Let $n$ be a real number such that $n \ne -1$.

Then:

$\map {\dfrac \d {\d x} } {\map u x^n} = n \map u x^{n - 1} \map {\dfrac \d {\d x} } {\map u x}$

## Proof

 $\ds \map {\dfrac \d {\d x} } {\map u x^n}$ $=$ $\ds \lim_{h \mathop \to 0} \frac {\paren {\map u {x + h} }^n - \paren {\map u x}^n} h$ $\ds$ $=$ $\ds \paren {\map u x}^n \lim_{h \mathop \to 0} \frac {\paren {\frac {\map u {x + h} } {\map u x} }^n - 1} h$ Power of Product $\ds$ $=$ $\ds \paren {\map u x}^n \lim_{h \mathop \to 0} \frac {\exp \paren {n \ln \frac {\map u {x + h} } {\map u x} } - 1} h$ Definition of Power to Real Number $\ds$ $=$ $\ds \paren {\map u x}^n \lim_{h \mathop \to 0} \paren {\frac {\map \exp {n \ln \frac {\map u {x + h} } {\map u x} } - 1} {n \ln \frac {\map u {x + h} } {\map u x} } } \paren {\frac {n \ln \frac {\map u {x + h} } {\map u x} } h}$ $\ds$ $=$ $\ds \paren {\map u x}^n \lim_{h \mathop \to 0} \frac {n \ln \frac {\map u {x + h} } {\map u x} } h$ Derivative of Exponential at Zero $\ds$ $=$ $\ds n \paren {\map u x}^n \lim_{h \mathop \to 0} \frac {\ln \frac {\map u {x + h} } {\map u x} } h$ $\ds$ $=$ $\ds n \paren {\map u x}^n \lim_{h \mathop \to 0} \frac {\map \ln {1 + \frac {\map u {x + h} - \map u x} {\map u x} } } h$ $\ds$ $=$ $\ds n \paren {\map u x}^n \lim_{h \mathop \to 0} \paren {\frac {\map \ln {1 + \frac {\map u {x + h} - \map u x} {\map u x} } } {\frac {\map u {x + h} - \map u x} {\map u x} } } \paren {\frac {\frac {\map u {x + h} - \map u x} {\map u x} } h }$ $\ds$ $=$ $\ds n \paren {\map u x}^n \lim_{h \mathop \to 0} \frac {\paren {\frac {\map u {x + h} - \map u x} {\map u x} } } h$ Derivative of Logarithm at One $\ds$ $=$ $\ds n \paren {\map u x}^n \lim_{h \mathop \to 0} \frac 1 {\map u x} \frac {\map u {x + h} - \map u x} h$ $\ds$ $=$ $\ds n \paren {\map u x}^{n - 1} \lim_{h \mathop \to 0} \frac {\map u {x + h} - \map u x} h$ Product of Powers $\ds$ $=$ $\ds n \paren {\map u x}^{n - 1} \map {\dfrac \d {\d x} } {\map u x}$

$\blacksquare$