Power rule

The power rule allows one to calculate the derivative of a power of a function in terms of the derivative of the function itself. The power rule states that

${\frac {d}{dx}}(x^{n})=nx^{n-1}$

for all integers Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n } . This rule is useful when combined with the chain rule. As an example we can compute the derivative of $f(x)=(\sin(x))^{n}$ as

$f'(x)={\frac {d}{dx}}(\sin(x))^{n}=n(\sin(x))^{n-1}\cos(x)$

Proof

The power rule is simple and elegant to prove with the definition of a derivative:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x) = \lim_{h\rarr0} \frac{f(x+h)-f(x)}{h}. }

Substituting Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = x^n } gives

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x) = \lim_{h\rarr0} \frac{(x+h)^n-x^n}{h}.}

The two polynomials in the numerator can be factored out. It is left as a series, since n can be any integer.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x) = \lim_{h\rarr0} \frac{((x+h) - x)( (x+h)^{n-1} + (x+h)^{n-2}x + (x+h)^{n-3}x^2 + \dots + (x+h)x^{n-2} + x^{n-1})}{h}. }

Then if the first factor is eliminated:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x) = \lim_{h\rarr0} \frac{(h)( (x+h)^{n-1} + (x+h)^{n-2}x + (x+h)^{n-3}x^2\dots + (x+h)x^{n-2} + x^{n-1})}{h}. }

Now, the "h" can be eliminated. This is important, since the denominator cannot go to zero':

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x) = \lim_{h\rarr0} (x+h)^{n-1} + (x+h)^{n-2}x + (x+h)^{n-3}x^2\dots + (x+h)x^{n-2} + x^{n-1}. }

With no "h" in the denominator, the limit can be evaluated by letting h=0.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x) = (x+0)^{n-1} + (x+0)^{n-2}x + (x+0)^{n-3}x^2\dots + (x+0)x^{n-2} + x^{n-1}. } :

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x) = x^{n-1} + x^{n-1} + x^{n-1}\dots + x^{n-1} + x^{n-1}. }

We see that there are n terms, so:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x) = nx^{n-1}. }

QED