How Do You Differentiate a Polynomial Function?

Apply the power rule to each term separately: multiply the coefficient by the exponent, then reduce the exponent by one. The derivative of every constant term is zero.

The Method

For a term of the form $ax^n$, the power rule is

$$\frac{d}{dx}\left(ax^n\right)=anx^{n-1}.$$

A polynomial is a sum or difference of individual terms, so differentiation can be applied term by term. This uses the sum and difference rules: the derivative of a sum or difference equals the sum or difference of the derivatives.

Use these steps:

1. Multiply each term's coefficient by its exponent.
2. Subtract one from that exponent.
3. Differentiate constants as zero.
4. Simplify the resulting expression.

The derivative represents the gradient of the polynomial's graph at each value of $x$. Differentiation lowers the degree of every non-constant term by one, so a polynomial of degree $m$ normally has a derivative of degree $m-1$.

Worked Example

Differentiate

$$f(x)=4x^5-3x^2+7x-9.$$

Applying the power rule to each term gives

$$f'(x)=4(5)x^{5-1}-3(2)x^{2-1}+7(1)x^{1-1}-0.$$

Therefore,

$$f'(x)=20x^4-6x+7.$$

Notice that $7x$ becomes $7$, because $x=x^1$ and $x^0=1$. The constant becomes zero because a constant does not change, so its gradient is zero.

For example, at $x=1$, $f'(1)=20-6+7=21$. This means the tangent to the graph of at has gradient . Substitution should be done only after finding the derivative function when the question asks for a gradient at a point. Writing this interpretation can earn reasoning credit in an applied question.

A common misconception is to subtract one from the coefficient as well as the exponent. This is incorrect: for $5x^3$, the derivative is $15x^2$, not $4x^2$.

IB Exam Technique

This is IB Mathematics: Analysis and Approaches SL 5.3, applicable to both Paper 1 and Paper 2. The command term Differentiate means “obtain the derivative of a function.” Show the power-rule step clearly, preserve negative signs, use notation such as (f'(x)) or $\frac{dy}{dx}$, and simplify your final answer.