Derivative Rules (Power, Product, Quotient)

Learn the power rule, product rule, quotient rule, and trig derivatives essential for AP Calculus AB differentiation problems.

Differentiation is the heart of AP Calculus AB. Once you move beyond limits, virtually every topic — tangent lines, rates of change, optimization, related rates, and even integration — relies on your ability to take derivatives quickly and accurately.

This guide covers the foundational derivative rules: the power rule, constant multiple rule, sum/difference rule, product rule, and quotient rule. We will also cover the derivatives of the six trigonometric functions. These rules are the building blocks you will use hundreds of times throughout the course, so fluency here is critical.

By the end of this guide, you should be able to differentiate any polynomial, rational, or trigonometric function without hesitation.

Core Concepts

The Definition of the Derivative

The derivative of ff at xx is defined as:

f(x)=limh0f(x+h)f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

This limit gives the instantaneous rate of change of ff at xx, and geometrically it represents the slope of the tangent line to the graph of ff at the point (x,f(x))(x, f(x)).

While the definition is important conceptually, the derivative rules below let you compute derivatives efficiently without returning to the limit every time.

Basic Differentiation Rules

Constant Rule: If f(x)=cf(x) = c (a constant), then f(x)=0f'(x) = 0.

Power Rule: If f(x)=xnf(x) = x^n for any real number nn, then:

f(x)=nxn1f'(x) = nx^{n-1}

This works for positive integers, negative integers, and fractions. Examples:

  • ddx[x5]=5x4\frac{d}{dx}[x^5] = 5x^4
  • ddx[x2]=2x3\frac{d}{dx}[x^{-2}] = -2x^{-3}
  • ddx[x]=ddx[x1/2]=12x1/2=12x\frac{d}{dx}[\sqrt{x}] = \frac{d}{dx}[x^{1/2}] = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}

Constant Multiple Rule: ddx[cf(x)]=cf(x)\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)

Sum/Difference Rule: ddx[f(x)±g(x)]=f(x)±g(x)\frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)

Together, these rules let you differentiate any polynomial term by term. For example:

ddx[3x47x2+5x9]=12x314x+5\frac{d}{dx}[3x^4 - 7x^2 + 5x - 9] = 12x^3 - 14x + 5

The Product Rule

When two functions are multiplied, you cannot simply differentiate each factor separately. The product rule states:

ddx[f(x)g(x)]=f(x)g(x)+f(x)g(x)\frac{d}{dx}[f(x) \cdot g(x)] = f'(x) \cdot g(x) + f(x) \cdot g'(x)

A helpful mnemonic: "derivative of the first times the second, plus the first times the derivative of the second."

Example: Find ddx[x3sinx]\frac{d}{dx}[x^3 \sin x].

=3x2sinx+x3cosx= 3x^2 \cdot \sin x + x^3 \cdot \cos x

The Quotient Rule

For a quotient of two functions:

ddx[f(x)g(x)]=f(x)g(x)f(x)g(x)[g(x)]2\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{[g(x)]^2}

A common mnemonic: "lo d-hi minus hi d-lo, over lo-lo" (where "lo" = denominator, "hi" = numerator, and "d" = derivative of).

Example: Find ddx[x2+1x3]\frac{d}{dx}\left[\frac{x^2 + 1}{x - 3}\right].

=2x(x3)(x2+1)(1)(x3)2=2x26xx21(x3)2=x26x1(x3)2= \frac{2x(x-3) - (x^2+1)(1)}{(x-3)^2} = \frac{2x^2 - 6x - x^2 - 1}{(x-3)^2} = \frac{x^2 - 6x - 1}{(x-3)^2}

Derivatives of Trigonometric Functions

You must memorize all six:

Function Derivative
sinx\sin x cosx\cos x
cosx\cos x sinx-\sin x
tanx\tan x sec2x\sec^2 x
cotx\cot x csc2x-\csc^2 x
secx\sec x secxtanx\sec x \tan x
cscx\csc x cscxcotx-\csc x \cot x

Notice the pattern: the "co-" functions (cosine, cotangent, cosecant) all have a negative sign in their derivatives.

Derivatives of Exponential and Logarithmic Functions

While covered more deeply elsewhere, you should know:

ddx[ex]=exddx[lnx]=1x\frac{d}{dx}[e^x] = e^x \qquad \frac{d}{dx}[\ln x] = \frac{1}{x}

ddx[ax]=axlnaddx[logax]=1xlna\frac{d}{dx}[a^x] = a^x \ln a \qquad \frac{d}{dx}[\log_a x] = \frac{1}{x \ln a}

When to Use Product Rule vs. Quotient Rule

Sometimes you have a choice. For example, x3x2\frac{x^3}{x^2} can be simplified to xx first — no quotient rule needed. Other times, rewriting a quotient as a product with a negative exponent lets you use the product rule instead:

f(x)g(x)=f(x)[g(x)]1\frac{f(x)}{g(x)} = f(x) \cdot [g(x)]^{-1}

The quotient rule and the product rule (with chain rule) will always give the same answer. Use whichever feels more natural for the given problem.

Strategy Tips

Tip 1: Simplify Before Differentiating

Always look for opportunities to simplify first. Expanding products, reducing fractions, or rewriting radicals as fractional powers can make differentiation much easier and reduce errors.

Tip 2: Rewrite Roots and Reciprocals as Powers

Convert x\sqrt{x} to x1/2x^{1/2}, 1x3\frac{1}{x^3} to x3x^{-3}, and x23\sqrt[3]{x^2} to x2/3x^{2/3}. This lets you apply the power rule directly.

Tip 3: Memorize Trig Derivatives Cold

You cannot afford to derive these from scratch during the exam. Flash-card drill until they are automatic. Remember the negative sign pattern for co-functions.

Tip 4: Check Your Work by Substituting a Value

After finding a derivative, plug in a specific xx-value and verify against a numerical approximation using the limit definition: f(a)f(a+0.001)f(a)0.001f'(a) \approx \frac{f(a+0.001) - f(a)}{0.001}.

Tip 5: Watch for Hidden Product/Quotient Structures

On the AP exam, functions are not always presented in obvious product or quotient form. Something like x2tanxx^2 \tan x requires the product rule, and sinx1+cosx\frac{\sin x}{1 + \cos x} requires the quotient rule.

Worked Example: Example 1

Problem

Find f(x)f'(x) if f(x)=4x52x3+x3f(x) = 4x^5 - \frac{2}{x^3} + \sqrt[3]{x}.

Solution

Rewrite using powers: f(x)=4x52x3+x1/3f(x) = 4x^5 - 2x^{-3} + x^{1/3}.

Apply the power rule term by term:

f(x)=20x42(3)x4+13x2/3=20x4+6x4+13x2/3f'(x) = 20x^4 - 2(-3)x^{-4} + \frac{1}{3}x^{-2/3} = 20x^4 + 6x^{-4} + \frac{1}{3}x^{-2/3}

Or equivalently: f(x)=20x4+6x4+13x23f'(x) = 20x^4 + \frac{6}{x^4} + \frac{1}{3\sqrt[3]{x^2}}.

Worked Example: Example 2

Problem

Find the derivative of g(x)=(3x2+1)(x42x)g(x) = (3x^2 + 1)(x^4 - 2x).

Solution

Use the product rule with f=3x2+1f = 3x^2 + 1 and h=x42xh = x^4 - 2x:

g(x)=6x(x42x)+(3x2+1)(4x32)g'(x) = 6x(x^4 - 2x) + (3x^2 + 1)(4x^3 - 2)

=6x512x2+12x56x2+4x32= 6x^5 - 12x^2 + 12x^5 - 6x^2 + 4x^3 - 2

=18x5+4x318x22= 18x^5 + 4x^3 - 18x^2 - 2

Worked Example: Example 3

Problem

Differentiate y=cosxx2y = \frac{\cos x}{x^2}.

Solution

Apply the quotient rule:

y=(sinx)(x2)(cosx)(2x)(x2)2=x2sinx2xcosxx4y' = \frac{(-\sin x)(x^2) - (\cos x)(2x)}{(x^2)^2} = \frac{-x^2 \sin x - 2x \cos x}{x^4}

Factor x-x from the numerator:

=x(xsinx+2cosx)x4=(xsinx+2cosx)x3= \frac{-x(x \sin x + 2 \cos x)}{x^4} = \frac{-(x \sin x + 2 \cos x)}{x^3}

Worked Example: Example 4

Problem

Find the equation of the tangent line to f(x)=x36x+2f(x) = x^3 - 6x + 2 at x=2x = 2.

Solution

First find the point: f(2)=812+2=2f(2) = 8 - 12 + 2 = -2. So the point is (2,2)(2, -2).

Find the slope: f(x)=3x26f'(x) = 3x^2 - 6, so f(2)=126=6f'(2) = 12 - 6 = 6.

Tangent line: y(2)=6(x2)y - (-2) = 6(x - 2), which gives y=6x14y = 6x - 14.

Worked Example: Example 5

Problem

Find ddx[x2secx]\frac{d}{dx}[x^2 \sec x].

Solution

Product rule with f=x2f = x^2 and g=secxg = \sec x:

ddx[x2secx]=2xsecx+x2secxtanx\frac{d}{dx}[x^2 \sec x] = 2x \sec x + x^2 \sec x \tan x

Factor: =xsecx(2+xtanx)= x \sec x(2 + x \tan x).

Practice Problems

  1. Problem 1

    Differentiate f(x)=7x63x2+5xf(x) = 7x^6 - 3x^{-2} + \frac{5}{\sqrt{x}}.

    Answer: f(x)=42x5+6x352x3/2f'(x) = 42x^5 + 6x^{-3} - \frac{5}{2}x^{-3/2}

    Problem 2

    Find ddx[(2x+1)(x3x)]\frac{d}{dx}[(2x + 1)(x^3 - x)].

    Answer: 2(x3x)+(2x+1)(3x21)=8x3+3x24x12(x^3 - x) + (2x+1)(3x^2 - 1) = 8x^3 + 3x^2 - 4x - 1

    Problem 3

    Differentiate y=x24x+5y = \frac{x^2 - 4}{x + 5}.

    Answer: y=x2+10x+4(x+5)2y' = \frac{x^2 + 10x + 4}{(x+5)^2}

    Problem 4

    Find f(π/4)f'(\pi/4) if f(x)=tanxsecxf(x) = \tan x - \sec x.

    Answer: f(x)=sec2xsecxtanxf'(x) = \sec^2 x - \sec x \tan x. At x=π/4x = \pi/4: f(π/4)=22f'(\pi/4) = 2 - \sqrt{2}

    Problem 5

    Find the slope of the tangent line to g(x)=xsinxg(x) = x \sin x at x=πx = \pi.

    Answer: g(x)=sinx+xcosxg'(x) = \sin x + x \cos x, so g(π)=0+π(1)=πg'(\pi) = 0 + \pi(-1) = -\pi

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Common Mistakes

  • Forgetting the power rule applies to negative and fractional exponents. Students often fail to rewrite 1x2\frac{1}{x^2} as x2x^{-2} and get stuck. Always rewrite before differentiating.
  • Applying the product rule when it's not needed. If f(x)=3x5f(x) = 3x^5, the 3 is a constant multiple — just use the constant multiple rule. The product rule works but is overkill.
  • Sign errors in the quotient rule. The numerator is fgfgf'g - fg', not fgfgfg' - f'g. The order matters! The derivative of the top comes first.
  • Forgetting the squared denominator. In the quotient rule, the denominator is [g(x)]2[g(x)]^2, not g(x)g(x).
  • Dropping the negative sign for co-function derivatives. ddx[cosx]=sinx\frac{d}{dx}[\cos x] = -\sin x, not sinx\sin x. The negative sign is the number one error with trig derivatives.
  • Not simplifying after differentiating. On free-response questions, a simplified answer is easier to work with and less prone to errors in subsequent steps.

Frequently Asked Questions

Do I need to know the limit definition of the derivative for the AP exam?

Yes. The AP exam occasionally tests the limit definition directly, asking you to recognize that limh0f(a+h)f(a)h\lim_{h \to 0} \frac{f(a+h) - f(a)}{h} equals f(a)f'(a) or to write the derivative using the definition. You should be comfortable with both the h0h \to 0 form and the xax \to a form.

When should I use the product rule versus just expanding?

If the product is easy to expand (e.g., two binomials), expanding first may be simpler. But if the product involves trig functions, exponentials, or complicated expressions, the product rule is essential. On the AP exam, many product rule problems involve functions that can't be expanded.

Is there a shortcut for the quotient rule?

Some students prefer to rewrite fg\frac{f}{g} as fg1f \cdot g^{-1} and use the product rule combined with the chain rule. This avoids memorizing a separate formula. Both methods give identical results.

How do I differentiate $y = \frac{5}{x^3}$? Product rule or power rule?

Neither the product rule nor quotient rule is needed. Rewrite as y=5x3y = 5x^{-3} and use the power rule: y=15x4=15x4y' = -15x^{-4} = -\frac{15}{x^4}.

Why is $\frac{d}{dx}[\cos x]$ negative but $\frac{d}{dx}[\sin x]$ positive?

This comes from the limit definition and the unit circle. When xx is small and positive, cosx\cos x is decreasing (the xx-coordinate of the point on the unit circle moves left), so its derivative is negative. Meanwhile, sinx\sin x is increasing for small positive xx.

Key Takeaways

  • The power rule is your workhorse. It handles polynomials, roots, and reciprocal powers — rewrite everything as xnx^n and apply nxn1nx^{n-1}.

  • Product rule: derivative of first times second, plus first times derivative of second. Practice this pattern until it's automatic.

  • Quotient rule: lo d-hi minus hi d-lo, over lo squared. Watch the sign order carefully — the subtraction is not commutative.

  • Memorize all six trig derivatives. Pay special attention to the negative signs on cosine, cotangent, and cosecant.

  • Simplify first when possible. Reducing fractions, expanding simple products, or rewriting with exponents often saves time and reduces errors.

  • Derivatives have geometric meaning. The derivative at a point gives the slope of the tangent line — connect algebraic computation to graphical interpretation.

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