Difference between revisions of "Power rule"

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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
Welcome to the guild TK
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<math> \frac{d}{dx} (x^n) = nx^{n-1} </math>
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for all integers <math> n </math>.  This rule is useful when combined with the [[chain rule]].  As an example we can compute the derivative of <math> f(x) = (\sin(x))^n</math> as
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<math> f'(x) = \frac{d}{dx} (\sin(x))^n = n(\sin(x))^{n-1}\cos(x) </math>
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==Proof==
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The power rule is simple and elegant to prove with the definition of a derivative:
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:<math>f'(x) = \lim_{h\rarr0} \frac{f(x+h)-f(x)}{h}. </math>
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Substituting <math> f(x) = x^n </math> gives
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:<math>f'(x) = \lim_{h\rarr0} \frac{(x+h)^n-x^n}{h}.</math>
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The two polynomials in the numerator can be factored out. It is left as a series, since n can be any integer.
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:<math>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}. </math>
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Then if the first factor is eliminated:
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:<math>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}. </math>
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Now, the "h" can be eliminated. This is important, since the denominator cannot go to zero':
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:<math>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}. </math>
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With no "h" in the denominator, the limit can be evaluated by letting h=0.
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:<math>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}. </math>:
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:<math>f'(x) = x^{n-1} + x^{n-1} + x^{n-1}\dots + x^{n-1} + x^{n-1}. </math>
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We see that there are ''n'' terms,  so:
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:<math>f'(x) = nx^{n-1}. </math>
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QED
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==Rules for finding derivatives==
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*Power rule
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*[[Constant-multiple rule]]
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*[[sum rule]]
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*[[Chain rule]]
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*[[Product rule]]
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*[[Quotient rule]]
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[[Category:Mathematics]]
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[[Category:Calculus]]
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[[Category:Differentiation]]

Latest revision as of 14:01, August 29, 2017

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

<math> \frac{d}{dx} (x^n) = nx^{n-1} </math>

for all integers <math> n </math>. This rule is useful when combined with the chain rule. As an example we can compute the derivative of <math> f(x) = (\sin(x))^n</math> as

<math> f'(x) = \frac{d}{dx} (\sin(x))^n = n(\sin(x))^{n-1}\cos(x) </math>

Proof

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

<math>f'(x) = \lim_{h\rarr0} \frac{f(x+h)-f(x)}{h}. </math>

Substituting <math> f(x) = x^n </math> gives

<math>f'(x) = \lim_{h\rarr0} \frac{(x+h)^n-x^n}{h}.</math>

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

<math>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}. </math>

Then if the first factor is eliminated:

<math>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}. </math>

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

<math>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}. </math>

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

<math>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}. </math>:
<math>f'(x) = x^{n-1} + x^{n-1} + x^{n-1}\dots + x^{n-1} + x^{n-1}. </math>

We see that there are n terms, so:

<math>f'(x) = nx^{n-1}. </math>

QED

Rules for finding derivatives