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Derivatives of Power Functions - Concept 15,954 views

Teacher/Instructor Norm Prokup
Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

The derivative of a power function involving x to the nth power (n being non-zero) can be derived using the definition of the derivative. The power function derivative is equal to x to the (n-1)th power times n. Many polynomial derivatives are based on derivatives of multiple power functions.

What what? I want to talk about the derivative of power of function. Now let's recall what a power of function is, a power of function is the one of the form f of x equals x to the n where n is some constant. Now let's find the derivative of a particular power of function f of x equals x cubed. Now recall the definition of the derivative f prime of x is the limit as h approaches zero of f of x plus h minus f of x over h. But first I want to focus on the difference quotient for this function. f of x plus h minus f of x over h is x cubed. When you plug in x+h you get x+h cubed minus f of x that's just x cubed all over h. So I've got to expand this binomial and if you recall your binomial expansions this is going to give me x cubed plus 3x squared h plus 3xh squared plus h cubed it's a big one minus x cubed all over h. Okay now we got a little bit of cancellation the x cubes go, and we're left with these 3 terms over h. So 3x squared h plus 3xh squared plus h cubed over h now we got more cancellation. Because a common factor of h in each of these terms and that'll cancel leaving 3x squared plus 3xh plus h squared.
Now this is my simplified difference quotient so I pop this back into my limit and I get the limit as h approaches zero of 3x squared plus 3xh plus h squared and as h goes to zero these two terms disappear and I'm left with 3x squared and that's my derivative. That's the derivative of f of x equals x cubed. Now I've got a little table here and it shows f of x equals x cubed and its derivative and you could do what I just did for f of x equals x to the fourth, x of the fifth and you'd find that the derivatives of those functions are 4x cubed and 5x of the fourth respectively.
And you may notice that there's a kind of a pattern here right, the n of the power comes down in front and it's replaced by an exponent that's 1 less. So if you have f of x equals x to the m its derivative is going to be n times x to the n minus 1, right? The original exponent comes down in front and it's x to the n minus 1, 1 less than the original exponent.