The derivative of f(x)=x2 was found to be f'(x)=2x. Here, the derivatives of higher powers of x shall be investigate to demonstrate a pattern. Note in the algebra shown below, Pascal's triangle is used to expand powers of (x+h)n.
The function f(x)=x3 is an antisymmetic function since f(x)=-f(-x), one can substitute x with some values to demonstrate this e.g. f(2)=8 and f(-2)=-8, therefore f(2)=-f(-2). A secant line passes through the points A(x,x3) and B(x+h,(x+h)3). f'(x) is found by taking the limit h → 0.
The function f(x)=x4 is a symmetic function since f(x)=f(-x), one can substitute x with some values to demonstrate this e.g. f(2)=16 and f(-2)=16, therefore f(2)=f(-2). A secant line passes through the points A(x,x4) and B(x+h,(x+h)4). f'(x) is found by taking the limit h → 0.
The function f(x)=x5 is an antisymmetic function since f(x)=-f(-x), one can substitute x with some values to demonstrate this e.g. f(2)=32 and f(-2)=-32, therefore f(2)=-f(-2). A secant line passes through the points A(x,x5) and B(x+h,(x+h)5). f'(x) is found by taking the limit h → 0.
The function f(x)=x-1 is an antisymmetic function since f(x)=-f(-x), one can substitute x with some values to demonstrate this e.g. f(2)=1/2 and f(-2)=-1/2, therefore f(2)=-f(-2). A secant line passes through the points A(x,1/x) and B(x+h,1/(x+h)). f'(x) is found by taking the limit h → 0.
The table summarizes our findings for the derivative of f(x)=xn for several integer n values. The results suggests that the derivative of f(x)=xn is f'(x)=nxn-1 for integer values of n. The Binomial expansion can be used to prove that the result holds for all positive integer values of n. However, the derivative rule is valid for all real values of n, including negative, fractional, and irrational values; the proof is beyond the scope of this page.
f(x) | f'(x) |
---|---|
x2 | 2x1 |
x3 | 3x2 |
x4 | 4x3 |
x5 | 5x4 |
x-1 | -x-2 |
xn | nxn-1 |