How do you take the derivative of a polynomial?

Category: Derivatives »

Course: Calculus »

Relevant MM Lessons:

Overview: Polynomial Derivatives

This question has its own dedicated lesson ». Make sure to check it out if you need a more thorough read, or want to become an expert on how polynomial derivatives show up on tests.Taking the derivative of polynomials is one of the fastest and most common operations you'll need to do in Calculus. The general formula (a.k.a. The Power Rule) for the derivative of any particular polynomial term looks like:$$\frac{d}{dx} \, x^n = nx^{n-1}$$And, using the linear operator property of derivatives on sums of terms, as outlined in the lesson on Derivative Rules », when we take a polynomial of a polynomial, we just go along and take the derivative of each term. Let's see a few examples.$$\frac{d}{dx} \, x^4 + x^6 - x^3$$According the Power Rule, the derivative of $x^4$ is $4x^3$, the derivative of $x^6$ is $6x^5$, and the derivative of $x^3$ is $3x^2$. Therefore, applying the derivative to the polynomial is the same as going along and finding the derivative of each term, and we get$$\frac{d}{dx} \, x^4 + x^6 - x^3 = 4x^3 + 6x^5 - 3x^2$$Not too bad, right?Next, remember that if you take the derivative of a term that already has a coefficient, that we can just keep the coefficient along for the ride, and take the derivative of the rest of the term. For example:$$\frac{d}{dx} \, 2x^6 = 2 \cdot \frac{d}{dx} \, x^6$$$$= 2 \left( 6x^5\right) = 12x^5$$Finally, we should remember that the derivative of $x$ is $1$, and the derivative of any constant $a$ is zero. Here's one last example with some of everything.$$\frac{d}{dx} \, 3x^4 + x^3 - 2x + 7$$$$\frac{d}{dx} \, 12x^3 + 3x^2 - 2$$
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