You CAN Ace Calculus

 basic derivative rules power rule Some of the practice problems require you to know the following rules also (in their basic form, not including the chain rule). If you don't know one or more of these rules, no worries. You can skip those problems and come back to them later. exponential derivative derivatives of trig functions You do NOT need to know the chain rule for anything on this page, including practice problems. For practice problems using the product rule and chain rule, see the chain rule page.

WikiBooks - Product Rule

### 17Calculus Subjects Listed Alphabetically

Single Variable Calculus

 Absolute Convergence Alternating Series Arc Length Area Under Curves Chain Rule Concavity Conics Conics in Polar Form Conditional Convergence Continuity & Discontinuities Convolution, Laplace Transforms Cosine/Sine Integration Critical Points Cylinder-Shell Method - Volume Integrals Definite Integrals Derivatives Differentials Direct Comparison Test Divergence (nth-Term) Test
 Ellipses (Rectangular Conics) Epsilon-Delta Limit Definition Exponential Derivatives Exponential Growth/Decay Finite Limits First Derivative First Derivative Test Formal Limit Definition Fourier Series Geometric Series Graphing Higher Order Derivatives Hyperbolas (Rectangular Conics) Hyperbolic Derivatives
 Implicit Differentiation Improper Integrals Indeterminate Forms Infinite Limits Infinite Series Infinite Series Table Infinite Series Study Techniques Infinite Series, Choosing a Test Infinite Series Exam Preparation Infinite Series Exam A Inflection Points Initial Value Problems, Laplace Transforms Integral Test Integrals Integration by Partial Fractions Integration By Parts Integration By Substitution Intermediate Value Theorem Interval of Convergence Inverse Function Derivatives Inverse Hyperbolic Derivatives Inverse Trig Derivatives
 Laplace Transforms L'Hôpital's Rule Limit Comparison Test Limits Linear Motion Logarithm Derivatives Logarithmic Differentiation Moments, Center of Mass Mean Value Theorem Normal Lines One-Sided Limits Optimization
 p-Series Parabolas (Rectangular Conics) Parabolas (Polar Conics) Parametric Equations Parametric Curves Parametric Surfaces Pinching Theorem Polar Coordinates Plane Regions, Describing Power Rule Power Series Product Rule
 Quotient Rule Radius of Convergence Ratio Test Related Rates Related Rates Areas Related Rates Distances Related Rates Volumes Remainder & Error Bounds Root Test Secant/Tangent Integration Second Derivative Second Derivative Test Shifting Theorems Sine/Cosine Integration Slope and Tangent Lines Square Wave Surface Area
 Tangent/Secant Integration Taylor/Maclaurin Series Telescoping Series Trig Derivatives Trig Integration Trig Limits Trig Substitution Unit Step Function Unit Impulse Function Volume Integrals Washer-Disc Method - Volume Integrals Work

Multi-Variable Calculus

 Acceleration Vector Arc Length (Vector Functions) Arc Length Function Arc Length Parameter Conservative Vector Fields Cross Product Curl Curvature Cylindrical Coordinates
 Directional Derivatives Divergence (Vector Fields) Divergence Theorem Dot Product Double Integrals - Area & Volume Double Integrals - Polar Coordinates Double Integrals - Rectangular Gradients Green's Theorem
 Lagrange Multipliers Line Integrals Partial Derivatives Partial Integrals Path Integrals Potential Functions Principal Unit Normal Vector
 Spherical Coordinates Stokes' Theorem Surface Integrals Tangent Planes Triple Integrals - Cylindrical Triple Integrals - Rectangular Triple Integrals - Spherical
 Unit Tangent Vector Unit Vectors Vector Fields Vectors Vector Functions Vector Functions Equations

Differential Equations

 Boundary Value Problems Bernoulli Equation Cauchy-Euler Equation Chebyshev's Equation Chemical Concentration Classify Differential Equations Differential Equations Euler's Method Exact Equations Existence and Uniqueness Exponential Growth/Decay
 First Order, Linear Fluids, Mixing Fourier Series Inhomogeneous ODE's Integrating Factors, Exact Integrating Factors, Linear Laplace Transforms, Solve Initial Value Problems Linear, First Order Linear, Second Order Linear Systems
 Partial Differential Equations Polynomial Coefficients Population Dynamics Projectile Motion Reduction of Order Resonance
 Second Order, Linear Separation of Variables Slope Fields Stability Substitution Undetermined Coefficients Variation of Parameters Vibration Wronskian

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17calculus > derivatives > product rule

Product Rule

The Product Rule is pretty straight-forward. If you have a function with two main parts that are multiplied together, for example $$h(x)=f(x) \cdot g(x)$$, the derivative is

$$\displaystyle{ \frac{dh}{dx} = }$$ $$\displaystyle{ \frac{d}{dx}[f \cdot g] = }$$ $$\displaystyle{ f'\cdot g + f \cdot g' }$$

An interesting thing to notice about the product rule is that the constant multiple rule is just a special case of the product rule. For example, if you have function $$f(x) = cg(x)$$, the product rule says $$f'(x) = (c)' g(x) + c g'(x) =$$ $$0 + c g'(x) = c g'(x)$$. Notice that we use the constant rule to say that $$d[c]/dx = 0$$.

And that's all you need to know to use the product rule. A common mistake many students make is to think that the product rule allows you to take the derivative of both terms and multiply them together. WRONG! If this confuses you, go back to the top of the page and reread the product rule and then go through some examples in your textbook.

Before you start using the product rule, it is important to know where it comes from. So take a few minutes to watch this video showing the proof of the product rule.

### PatrickJMT - Product Rule Proof [6min-6secs]

video by PatrickJMT

 Okay, practice problem time. Once you are finished with those, the quotient rule is the next logical step. quotient rule →

### Practice

Instructions - - Unless otherwise instructed, calculate the derivatives of the following functions using the product rule, giving your final answers in simplified, factored form.

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. Here is a list converting the old numbering system to the new.

Derivative Product Rule - Practice Problems Conversion

[A01-938] - [A02-1928] - [A03-939] - [A04-925] - [A05-927] - [A06-1311] - [A07-928] - [A08-929] - [A09-930]

[A10-931] - [A11-932] - [B01-926] - [B02-935] - [B03-936] - [B04-940] - [B05-937] - [B06-933] - [C01-934]

Please update your notes to this new numbering system. The display of this conversion information is temporary.

GOT IT. THANKS!

Basic Problems

Here are some problems that use only the product rule, the power rule and the other basic rules on the main derivatives page.

$$f(x) = x(25-x)$$

Problem Statement

Calculate the derivative of $$f(x) = x(25-x)$$ using the product rule.

Solution

### 938 solution video

video by Krista King Math

$$f(x) = (x-2)(x+3)$$

Problem Statement

Calculate the derivative of $$f(x) = (x-2)(x+3)$$ using the product rule.

$$f'(x)=2x+1$$

Problem Statement

Calculate the derivative of $$f(x) = (x-2)(x+3)$$ using the product rule.

Solution

### 1928 solution video

video by PatrickJMT

$$f'(x)=2x+1$$

$$f(x) = (2x+3)(3x-2)$$

Problem Statement

Calculate the derivative of $$f(x) = (2x+3)(3x-2)$$ using the product rule.

Solution

### 939 solution video

video by Krista King Math

$$h(x)=(x^2+5x+7)(x^3+2x-4)$$

Problem Statement

Calculate the derivative of $$h(x) = (x^2+5x+7)(x^3+2x-4)$$ using the product rule.

$$h'(x)=5x^4+20x^3+27x^2+12x-6$$

Problem Statement

Calculate the derivative of $$h(x) = (x^2+5x+7)(x^3+2x-4)$$ using the product rule.

Solution

 Use the product rule. $$\displaystyle{ h'(x) = (x^2 + 5x + 7)\frac{d}{dx}[x^3 + 2x - 4] + (x^3 + 2x - 4) \frac{d}{dx}[x^2 + 5x + 7] }$$ Take the derivative of the appropriate terms. $$h'(x) = (x^2 + 5x + 7)(3x^2 + 2) + (x^3 + 2x - 4)(2x + 5)$$ Check for common terms. In this case, there are not any, so multiply out. Leaving it in this form is not usually considered proper form but as usual check with your instructor to see what they expect. $$h'(x) = (x^2)(3x^2 + 2) + 5x(3x^2 + 2) + 7(3x^2 + 2) + (x^3)(2x+5) + 2x(2x+5) - 4(2x+5)$$ $$h'(x) = 3x^4 + 2x^2 + 15x^3 + 10x + 21x^2 + 14 + 2x^4 + 5x^3 + 4x^2 + 10x - 8x - 20$$ Collect common terms. $$h'(x) = 5x^4 + 20x^3 + 27x^2 + 12x - 6$$

$$h'(x)=5x^4+20x^3+27x^2+12x-6$$

$$g(x) = (x^3-7x^2+4)$$ $$(3x^2+14)$$

Problem Statement

Calculate the derivative of $$g(x) = (x^3-7x^2+4)$$ $$(3x^2+14)$$ using the product rule.

$$g'(x) = x(15x^3-84x^2+42x-172)$$

Problem Statement

Calculate the derivative of $$g(x) = (x^3-7x^2+4)$$ $$(3x^2+14)$$ using the product rule.

Solution

 First, we apply the product rule. $$\displaystyle{ \frac{dg}{dx} = }$$ $$\displaystyle{ (x^3-7x^2+4)\cdot }$$ $$\displaystyle{ \frac{d}{dx}[3x^2+14]+ }$$ $$\displaystyle{ (3x^2+14)\cdot }$$ $$\displaystyle{ \frac{d}{dx}[x^3-7x^2+4] }$$ Now, we evaluate the derivatives in the above equation. $$\displaystyle{ \frac{dg}{dx} = }$$ $$(x^3-7x^2+4)(6x)$$ $$+$$ $$(3x^2+14)(3x^2-14x)$$ Before we simplify by multiplying out, let's look for common factors that we can factor out. Notice that there is an $$x$$ in both of the terms. So we factor that out. $$g'(x) = x[6x^3-42x^2+24+(3x^2+14)(3x-14)]$$ Now, we multiply out the terms inside the brackets. $$g'(x) = x(15x^3-84x^2+42x-172)$$

Factoring right after taking the derivative, before multiplying out, makes it easier to see common terms.

$$g'(x) = x(15x^3-84x^2+42x-172)$$

$$g(x) = (3x^4+2x-1)$$ $$(x^5-2x^2)$$

Problem Statement

Calculate the derivative of $$g(x) = (3x^4+2x-1)$$ $$(x^5-2x^2)$$ using the product rule.

Solution

### 1311 solution video

video by PatrickJMT

These problems require you to know how to take the derivative of exponential functions.

$$\displaystyle{f(x)=x^3e^x}$$

Problem Statement

Calculate the derivative of $$\displaystyle{f(x)=x^3e^x}$$ using the product rule.

$$\displaystyle{\frac{d}{dx}\left[x^3e^x\right]=x^2e^x(x+3)}$$

Problem Statement

Calculate the derivative of $$\displaystyle{f(x)=x^3e^x}$$ using the product rule.

Solution

 $$\displaystyle{ \frac{d}{dx}\left[x^3e^x\right] }$$ Use the product rule. $$\displaystyle{ x^3\cdot\frac{d[e^x]}{dx}+e^x\cdot\frac{d[x^3]}{dx} }$$ Take the derivative of the two terms. $$x^3\cdot e^x+e^x\cdot3x^2$$ Simplify by factoring. $$x^2e^x(x+3)$$

$$\displaystyle{\frac{d}{dx}\left[x^3e^x\right]=x^2e^x(x+3)}$$

$$\displaystyle{y=7e^{2x}}$$

Problem Statement

Calculate the derivative of $$\displaystyle{y=7e^{2x}}$$ using the product rule.

$$\displaystyle{\frac{d}{dx}\left[7e^{2x}\right]=14e^{2x}}$$

Problem Statement

Calculate the derivative of $$\displaystyle{y=7e^{2x}}$$ using the product rule.

Solution

 $$\displaystyle{ \frac{d}{dx} \left[ 7e^{2x} \right] }$$ $$\displaystyle{ 7 \frac{d}{dx}[ e^x \cdot e^x ] }$$ $$\displaystyle{ 7e^x \frac{d[e^x]}{dx} + 7e^x \frac{d[e^x]}{dx} }$$ $$7e^x (e^x) + 7e^x (e^x)$$ $$14e^{2x}$$

Note: The derivative is more easily calculated using the chain rule. However, since you are not required to know the chain rule for this page, this is a valid way to work it without the chain rule.

$$\displaystyle{\frac{d}{dx}\left[7e^{2x}\right]=14e^{2x}}$$

$$\displaystyle{e^x\sqrt{x}}$$

Problem Statement

Calculate the derivative of $$\displaystyle{e^x\sqrt{x}}$$ using the product rule.

$$\displaystyle{ \frac{e^x (1 + 2x)}{2x^{1/2}} }$$

Problem Statement

Calculate the derivative of $$\displaystyle{e^x\sqrt{x}}$$ using the product rule.

Solution

 $$\displaystyle{ \frac{d}{dx}\left[ e^x \sqrt{x} \right] }$$ Rewrite $$\sqrt{x}$$ as $$x^{1/2}$$ so that we can use the power rule. $$\displaystyle{ \frac{d}{dx}\left[ e^x \left( x^{1/2} \right) \right] }$$ Use the product rule. $$\displaystyle{ (e^x) \frac{d[x^{1/2}]}{dx} + \left( x^{1/2} \right) \frac{d[e^x]}{dx} }$$ $$e^x (1/2)x^{-1/2} + x^{1/2} e^x$$ $$\displaystyle{ \left[ \frac{e^x}{2x^{1/2}} \right] (1 + 2x) }$$

Notes -
After we used the product rule, we just used algebra to simplify and factor. Although there are many ways to write the final answer, we usually want all factors written with positive exponents, except possibly exponential terms. Your instructor may or may not enforce this but most textbooks will write it this way. Writing your answer as $$e^x \left[ (1/2)x^{-1/2} + x^{1/2} \right]$$ is not considered completely factored because both factors inside the parentheses have an x term.

$$\displaystyle{ \frac{e^x (1 + 2x)}{2x^{1/2}} }$$

These problems require you to know how to take the derivative of trig functions.

$$\displaystyle{x^4\tan(x)}$$

Problem Statement

Calculate the derivative of $$\displaystyle{x^4\tan(x)}$$ using the product rule.

$$\displaystyle{ x^3[x \sec^2(x)+4 \tan(x)] }$$

Problem Statement

Calculate the derivative of $$\displaystyle{x^4\tan(x)}$$ using the product rule.

Solution

$$\displaystyle{\frac{d}{dx}[ x^4 \tan(x) ] = x^4 \sec^2(x) + 4x^3 \tan(x) = }$$ $$\displaystyle{x^3[x \sec^2(x)+4 \tan(x)]}$$

Note: Not factoring out the $$x^3$$ term, may cost you points. Check with your instructor to see what they expect.

$$\displaystyle{ x^3[x \sec^2(x)+4 \tan(x)] }$$

$$\displaystyle{f(x)=\cos(x)\sin(x)}$$

Problem Statement

Calculate the derivative of $$\displaystyle{f(x)=\cos(x)\sin(x)}$$ using the product rule.

$$f'(x) = \cos^2(x) - \sin^2(x)$$

Problem Statement

Calculate the derivative of $$\displaystyle{f(x)=\cos(x)\sin(x)}$$ using the product rule.

Solution

 $$\displaystyle{ f'(x) = \frac{d}{dx}\left[ \cos(x) \sin(x) \right] }$$ $$\displaystyle{ f'(x) = \cos(x) \frac{d}{dx}[ \sin(x)] + \sin(x) \frac{d}{dx}[ \cos(x)] }$$ $$f'(x) = \cos(x)[ \cos(x)] + \sin(x)[-\sin(x)]$$ $$f'(x) = \cos^2(x) - \sin^2(x)$$

$$f'(x) = \cos^2(x) - \sin^2(x)$$

Intermediate Problems

$$\displaystyle{g(x)=(x^2+3x-5)\sqrt[3]{x}}$$

Problem Statement

Calculate the derivative of $$\displaystyle{g(x)=(x^2+3x-5)\sqrt[3]{x}}$$ using the product rule.

$$\displaystyle{\frac{d}{dx}\left[(x^2+3x-5)\sqrt[3]{x}\right]=\frac{7x^2+12x-5}{3x^{2/3}}}$$

Problem Statement

Calculate the derivative of $$\displaystyle{g(x)=(x^2+3x-5)\sqrt[3]{x}}$$ using the product rule.

Solution

 Rewrite $$\sqrt[3]{x}$$ as $$x^{1/3}$$ to use the power rule. $$\displaystyle{ g'(x) = \frac{d}{dx}\left[ (x^2+3x-5)(x^{1/3}) \right] }$$ Use the product rule. $$\displaystyle{ g'(x) = (x^2 + 3x - 5) \cdot \frac{d[x^{1/3}]}{dx} + \left( x^{1/3} \right) \cdot \frac{d[x^2+3x-5]}{dx} }$$ Take the two derivatives. $$g'(x) = (x^2 + 3x - 5) \cdot (1/3)x^{-2/3} + x^{1/3} \cdot (2x+3)$$ $$\displaystyle{ g'(x) = \frac{ x^2 + 3x - 5}{ 3x^{2/3}} + \left[ x^{1/3} \cdot (2x+3) \right] \left[\frac{3x^{2/3}}{3x^{2/3}} \right] }$$ $$\displaystyle{ g'(x) = \frac{ x^2 + 3x - 5 + (3x)(2x+3) }{3x^{2/3}} }$$ $$\displaystyle{ g'(x) = \frac{7x^2 + 12x - 5}{3x^{2/3}} }$$

Notes -
After we used the product rule, we just used algebra to simplify and factor. Although there are many ways to write the final answer, we usually want all factors written with positive exponents. Your instructor may or may not enforce this but most textbooks will write it this way.
Also, you may see $$x^{2/3}$$ written as $$\sqrt[3]{x^2}$$. Although this is correct notation, in calculus we usually leave our answer in fractional exponent form. Ask your instructor which form he/she prefers or check your textbook to see which form it shows the most.

$$\displaystyle{\frac{d}{dx}\left[(x^2+3x-5)\sqrt[3]{x}\right]=\frac{7x^2+12x-5}{3x^{2/3}}}$$

$$\displaystyle{f(x)=\left[\frac{1}{x^2}-\frac{3}{x^4}\right]\cdot}$$ $$(x+5x^3)$$

Problem Statement

Calculate the derivative of $$\displaystyle{f(x)=\left[\frac{1}{x^2}-\frac{3}{x^4}\right]\cdot}$$ $$(x+5x^3)$$ using the product rule.

Solution

### 935 solution video

video by Krista King Math

$$H(u)=(u-\sqrt{u})$$ $$(u+\sqrt{u})$$

Problem Statement

Calculate the derivative of $$H(u)=(u-\sqrt{u})$$ $$(u+\sqrt{u})$$ using the product rule.

Solution

### 936 solution video

video by Krista King Math

$$f(x)=x(2x-1)$$ $$(2x+1)$$

Problem Statement

Calculate the derivative of $$f(x)=x(2x-1)$$ $$(2x+1)$$ using the product rule.

Solution

### 940 solution video

video by Krista King Math

$$f(x)=(2x+1)$$ $$(4-x^2)(1+x^2)$$

Problem Statement

Calculate the derivative of $$f(x)=(2x+1)$$ $$(4-x^2)(1+x^2)$$ using the product rule.

Solution

### 937 solution video

video by Krista King Math

Here is one with a trig derivative.

$$\displaystyle{\frac{3}{x}\cot(x)}$$

Problem Statement

Calculate the derivative of $$\displaystyle{\frac{3}{x}\cot(x)}$$ using the product rule.

$$\displaystyle{ \frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right] = \frac{-3}{x^2}\left[ x \csc^2(x) + \cot(x) \right] }$$

Problem Statement

Calculate the derivative of $$\displaystyle{\frac{3}{x}\cot(x)}$$ using the product rule.

Solution

 $$\displaystyle{ \frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right] }$$ $$\displaystyle{ \frac{3}{x}\frac{d}{dx}[\cot(x)] + \cot(x) \frac{d}{dx}[3/x] }$$ $$\displaystyle{ \frac{3}{x} (-\csc^2(x)) + \cot(x) \frac{d}{dx}[3x^{-1}] }$$ $$\displaystyle{ \frac{-3}{x} \csc^2(x) + \cot(x) [3(-1)x^{-2}] }$$ $$\displaystyle{ \frac{-3}{x} \csc^2(x) - \frac{3}{x^2} \cot(x) }$$ $$\displaystyle{ \frac{-3}{x^2}[ x \csc^2(x) + \cot(x) ] }$$

$$\displaystyle{ \frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right] = \frac{-3}{x^2}\left[ x \csc^2(x) + \cot(x) \right] }$$

Here is a problem that has both trig and exponential derivatives. Give it a try.

$$f(x)=4x^2e^x$$ $$\sin(x)\sec(x)$$

Problem Statement

Calculate the derivative of $$f(x)=4x^2e^x$$ $$\sin(x)\sec(x)$$ using the product rule.

Solution

### 934 solution video

video by Krista King Math