# Derivatives of Polynomials

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

## Formula: Constant Multiple Rule in Differentiation

The coefficient *k* is not affected

by the differentiation.

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

This formula is needed

to solve the derivatives of polynomials.

## Formula: Sum Rule (and Subtraction Rule) in Differentiation

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

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*)]' = *k**f*'(*x*).

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

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.

[*x*^{2}]' = 2*x*^{1}

[*x*]' = 1*x*^{0}

[+1]' = 0

Power rule in differentiation (Part 1)

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

## Example 2

Focus on the variable parts

and differentiate each term.

[*x*^{5}]' = 5*x*^{4}

[*x*^{4}]' = 4*x*^{3}

[*x*^{2}]' = 2*x*^{1}

[*x*]' = 1*x*^{0}

[+14]' = 0

Power rule in differentiation (Part 1)

So *f*'(*x*) = 5*x*^{4} - 32*x*^{3} + 14*x* - 3.