# How do you use the definition of derivative in terms of limits, prove that the derivative of #x^n = nx^(n-1)#?

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Using the alternative definition of derivative:

So:

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To prove that the derivative of (x^n) is (nx^{n-1}) using the definition of derivative in terms of limits, we start with the definition of the derivative:

[ f'(x) = \lim_{{h \to 0}} \frac{{f(x + h) - f(x)}}{h} ]

For the function (f(x) = x^n), we substitute it into the definition:

[ f'(x) = \lim_{{h \to 0}} \frac{{(x + h)^n - x^n}}{h} ]

Expanding ((x + h)^n) using the binomial theorem:

[ (x + h)^n = x^n + nx^{n-1}h + \frac{{n(n-1)}}{2}x^{n-2}h^2 + \ldots ]

Substitute the expansion into the definition of the derivative:

[ f'(x) = \lim_{{h \to 0}} \frac{{(x^n + nx^{n-1}h + \ldots - x^n)}}{h} ]

Simplify the expression:

[ f'(x) = \lim_{{h \to 0}} \frac{{nx^{n-1}h + \ldots}}{h} ]

Cancel out the (h) terms:

[ f'(x) = \lim_{{h \to 0}} nx^{n-1} + \ldots ]

As (h) approaches 0, all terms involving (h) vanish except (nx^{n-1}), yielding:

[ f'(x) = nx^{n-1} ]

Therefore, the derivative of (x^n) is (nx^{n-1}).

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