What is the power rule for taking derivatives of polynomial functions?

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The power rule for taking derivatives of polynomial functions states that the derivative of x raised to the power of n is nx^(n-1). This rule applies specifically to functions of the form x^n, where x is the variable and n is a constant. The power rule can be effectively combined with the sum and constant factor rules of differentiation to simplify the process of finding derivatives. A proof using the limit definition of a derivative confirms this rule, demonstrating the cancellation of terms and the application of the Binomial Theorem. Overall, the power rule is a fundamental tool in calculus for differentiating polynomial expressions.
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Definition/Summary

A method used to take the derivative of a polynomial function.

Equations

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

Extended explanation

Power rule applies to a function of the form x^{n}, where x is the variable and n is a constant. Used in combination with the sum and constant factor rules of differentiation, power rule can be a powerful tool for taking derivatives.

Proof:

We can apply the limit definition of a derivative to this specific function:
f'(x) := \lim_{h→0} \frac{f(x+h)-f(x)}{h}
Substituting in gives us:
\frac{d}{dx} x^{n} = \lim_{h→0} \frac{(x+h)^{n}-x^{n}}{h}
If we then expand using Binomial Theorem:
\frac{d}{dx} x^{n} = \lim_{h→0} \frac{x^{n}+nx^{n-1}h+\binom{n}{2}x^{n-2}h^{2}+\cdots+h^{n} -x^{n}}{h}
We can then cancel the first and last x^{n} terms and distribute the h from the denominator:
\frac{d}{dx} x^{n} = \lim_{h→0} nx^{n-1}+\binom{n}{2}x^{n-2}h+\cdots+h^{n-1}
Finally, we take the limit by substituting in h=0:
\frac{d}{dx} x^{n} = nx^{n-1}+\binom{n}{2}x^{n-2}0+\cdots+0^{n-1}
\frac{d}{dx} x^{n} = nx^{n-1}

Example 1:
f(x) = x^{189}
f'(x) = 189x^{189-1} = 189x^{188}
Example 2:
f(x) = 3x^{3}+7x^{2}+8x+2
f'(x) = 9x^{2}+14x+8
Example 3:
f(x) = 3\sqrt{x}
f'(x) = 3×1/2\ x^{(1/2-1)} = \frac{3}{2\sqrt{x}}

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The power rule is a consecutive application of the product or Leibniz rule:
$$
\dfrac{d}{dx}x^n = \dfrac{d}{dx}(x\cdot x^{n-1})=\left( \dfrac{d}{dx} x\right) \cdot x^{n-1} + x\cdot \dfrac{d}{dx}x^{n-1}=1\cdot x^{n-1} + x\cdot (n-1)\cdot x^{n-2}=n \cdot x^{n-1}
$$
by induction.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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