Derivatives of Power Functions - Concept
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.