Derivatives of Polynomials

Derivatives of Polynomials

How to find the derivative of a polynomial: formulas, proofs, examples, and their solutions.

Formula: Constant Multiple Rule in Differentiation

[kf(x)]' = k*f'(x)

The coefficient k is not affected
by the differentiation.

So [kf(x)]' = kf'(x).

This formula is needed
to solve the derivatives of polynomials.

Formula: Sum Rule (and Subtraction Rule) in Differentiation

[f(x) +- g(x)]' = f'(x) +- g'(x)

The plus minus signs (±) are not affected
by the differentiation.

So [f(x) ± g(x)]' = f'(x) ± g'(x).

This formula is also needed
to solve the derivatives of polynomials.

Proofs

Constant Multiple Rule in Differentiation: Proof of the Formula

Use the definition of the derivative function.

Take the coefficient k out by factoring.

Take the coefficient k out from the limit.

The limit part is f'(x).

So [kf(x)]' = kf'(x).

Sum Rule (and Subtraction Rule) in Differentiation: Proof of the Formula

Use the definition of the derivative function.

Rewrite the terms like this:
f(x + h) ± g(x + h) - f(x) ± [-g(x)].

Combine the gray terms:
[f(x + h) - f(x)].

And combine the dark gray terms:
±[g(x + h) - g(x)].

Split the fraction into two parts.

Then the limit of the former fraction is f'(x).
And the limit of the latter fraction is g'(x).

So [f(x) ± g(x)]' = f'(x) ± g'(x).

Example 1

For the given f(x), find f'(x). f(x) = x^2 - 3x + 1

As you've seen above,
the coefficients and the plus minus signs
are not affected by the differentiation.

So focus on the variable parts
and differentiate each term.

[x2]' = 2x1
[x]' = 1x0
[+1]' = 0

Power rule in differentiation (Part 1)

So f'(x) = 2x - 3.

Example 2

For the given f(x), find f'(x). f(x) = x^5 - 8x^4 + 7x^2 - 3x + 14

Focus on the variable parts
and differentiate each term.

[x5]' = 5x4
[x4]' = 4x3
[x2]' = 2x1
[x]' = 1x0
[+14]' = 0

Power rule in differentiation (Part 1)

So f'(x) = 5x4 - 32x3 + 14x - 3.