﻿ How to Use the Product Rule for Derivatives - 16 Practice Problems explained step by step with interactive problems, showing all work.

# How to Use the Product Rule for Derivatives: Practice Problems

Step 1

Identify the factors in the function.

$$f(x) = \blue{5x^8}\red{\sin 4x}$$

Step 2

Differentiate using the product rule.

\begin{align*} f'(x) & = \blue{40x^7}\sin 4x + 5x^8\red{(4\cos 4x)}\\[6pt] & = 40x^7\sin 4x + 20x^8\cos 4x \end{align*}

$$f'(x) = 40x^7\sin 4x + 20x^8\cos 4x$$

Step 1

Identify the factors in the function.

$$f(x) = \blue{\sqrt x}\,\red{\sin \pi x}$$

Step 2

Rewrite the square-root as a power function.

$$f(x) = \blue{x^{1/2}}\,\red{\sin \pi x}$$

Step 3

Differentiate using the product rule

\begin{align*} f'(x) & = \blue{\frac 1 2 x^{-1/2}}\sin \pi x + x^{1/2}\red{(\pi\cos\pi x)}\\[6pt] & = \frac 1 2 x^{-1/2}\sin \pi x + \pi x^{1/2}\cos\pi x \end{align*}

Step 4

(Optional) Rewrite the derivative using radicals.

\begin{align*} f'(x) & = \frac 1 2 x^{-1/2}\sin \pi x + \pi x^{1/2}\cos\pi x\\[6pt] & = \frac 1 2 \cdot \frac 1 {x^{1/2}}\sin \pi x + \pi \sqrt x\,\cos\pi x\\[6pt] & = \frac 1 {2\sqrt x}\sin \pi x + \pi \sqrt x\,\cos\pi x \end{align*}

Step 5

Evaluate $$f'\left(\frac 1 4\right)$$.

\begin{align*} f'\left(\frac 1 4\right) & = \frac 1 {2\sqrt{1/4}}\sin \frac \pi 4 + \pi \sqrt{1/4}\,\cos\frac \pi 4\\[6pt] & = \frac 1 {2(1/2)}\left(\frac{\sqrt 2} 2\right) + \pi (1/2)\left(\frac{\sqrt 2} 2\right)\\[6pt] & = \frac{\sqrt 2} 2 + \frac \pi 2\left(\frac{\sqrt 2} 2\right)\\[6pt] & = \frac{\sqrt 2} 2 + \frac{\pi\sqrt 2} 4\\[6pt] & = \frac{2\sqrt 2 + \pi\sqrt 2} 4\\[6pt] & = \frac{\sqrt 2(2 + \pi)} 4 \end{align*}

$$\displaystyle f'\left(\frac 1 4\right) = \frac{\sqrt 2(2 + \pi)} 4 \approx 1.8178$$

Step 1

Identify the factors in the function.

$$f(x) = \blue{\frac 2 3 x^{-3/7}}\red{\cos x}$$

Step 2

Differentiate using the product rule.

\begin{align*} f'(x) & = \blue{\frac 2 3 \left(-\frac 3 7\right)x^{-3/7 -1}}\cos x + \frac 2 3 x^{-3/7}\red{(-\sin x)}\\[6pt] & = -\frac 2 7x^{-10/7}\cos x - \frac 2 3 x^{-3/7}\sin x \end{align*}

$$\displaystyle f'(x) = -\frac 2 7x^{-10/7}\cos x - \frac 2 3 x^{-3/7}\sin x$$

Step 1

Identify the factors in the function.

$$f(x) = \blue{-9x^{11}}\red{\cos 2x}$$

Step 2

Differentiate using the product rule.

\begin{align*} f'(x) & = \blue{-9\left(11x^{10}\right)}\cos 2x + (-9x^{11})\red{(-2\sin 2x)}\\[6pt] & = -99x^{10}\cos 2x + 18x^{11}\sin 2x \end{align*}

Step 3

Evaluate $$f'(1)$$.

$$f'(1) = -99(1)^{10}\cos 2(1) + 18(1)^{11}\sin 2(1) = -99\cos 2 + 18\sin 2\\ \approx 57.5659$$

$$f'(1) = -99\cos 2 + 18\sin 2 \approx 57.5659$$

Step 1

Identify the factors in the function.

$$f(x) = \blue x\,\red{e^{2x}}$$

Step 2

Differentiate using the product rule.

\begin{align*} f'(x) & = \blue 1\cdot e^{2x} + x\red{(2e^{2x})}\\[6pt] & = e^{2x} + 2xe^{2x} \end{align*}

Step 3

(Optional) Factor the derivative.

$$f'(x) = \blue{e^{2x}} + 2x\blue{e^{2x}} = (1+2x)\blue{e^{2x}}$$

$$(f'(x) = (1+2x)e^{2x}$$

Step 1

Identify the factors in the function.

$$f(x) = \blue{x^5}\,\red{e^{-0.4x}}$$

Step 2

Differentiate using the power rule.

\begin{align*} f'(x) & = \blue{5x^4}\,e^{-0.4x} + x^5\red{(-0.4e^{-0.4x})}\\[6pt] & = 5x^4\,e^{-0.4x} -0.4x^5\,e^{-0.4x} \end{align*}

Step 3

(Optional) Factor the derivative.

\begin{align*} f'(x) & = 5x^4\,e^{-0.4x} -0.4x^5\,e^{-0.4x}\\[6pt] & = 5x^4\,e^{-0.4x} -\frac 4 {10}x^5\,e^{-0.4x}\\[6pt] & = 5x^4\blue{e^{-0.4x}} -\frac 2 5x^5\blue{e^{-0.4x}}\\[6pt] & = \blue{e^{-0.4x}}\left(5\red{x^4} -\frac 2 5\red{x^5}\right)\\[6pt] & = \red{x^4}e^{-0.4x}\left(5 -\frac 2 5x\right)\\[6pt] & = \frac 1 5 x^4e^{-0.4x}\left(25 -2x\right)\\[6pt] & = \frac 1 5 x^4(25 -2x)e^{-0.4x} \end{align*}

Step 4

Evaluate $$f'(0)$$.

$$f'(0) = 1 5 (0)^4(25 -2(0))e^{-0.4(0)} = 0$$

$$\displaystyle f'(0) = 0$$

Step 1

Identify the factors in the function.

$$f(x) = \blue{\sin 5x}\red{\cos 2x}$$

Step 2

Differentiate using the product rule.

\begin{align*} f'(x) & = \blue{5\cos 5x}\cos 2x + \sin 5x \red{(-2\sin 2x)}\\[6pt] & = 5\cos 5x \cos 2x - 2\sin 5x \sin 2x \end{align*}

$$f'(x) = 5\cos 5x \cos 2x - 2\sin 5x \sin 2x$$

Step 1

Identify the factors in the function.

$$f(x) = \blue{\sin\left(\frac \pi 2 x\right)}\red{\cos\left(\frac \pi 3 x\right)}$$

Step 2

Differentiate using the product rule.

\begin{align*} f'(x) & = \blue{\frac \pi 2\cos\left(\frac \pi 2 x\right)}\cos\left(\frac\pi 3 x\right) + \sin\left(\frac \pi 2 x\right)\red{\left[-\frac \pi 3 \sin\left(\frac \pi 3 x\right)\right]}\\[6pt] & = \frac \pi 2\cos\left(\frac \pi 2 x\right)\cos\left(\frac\pi 3 x\right) -\frac \pi 3 \sin\left(\frac \pi 2 x\right)\sin\left(\frac \pi 3 x\right) \end{align*}

Step 3

Evaluate the derivative at $$x = 4$$.

\begin{align*} f'(\blue 4) & = \frac \pi 2\cos\left(\frac{\blue 4\pi} 2\right)\cos\left(\frac{\blue 4 \pi} 3\right) -\frac \pi 3 \sin\left(\frac{\blue 4 \pi} 2\right)\sin\left(\frac{\blue 4 \pi} 3\right)\\[6pt] & = \frac \pi 2\cos(2\pi)\cos\left(\frac{4\pi} 3\right) -\frac \pi 3 \sin(2\pi)\sin\left(\frac{4 \pi} 3\right)\\[6pt] & = \frac \pi 2(1)\left(-\frac 1 2\right) -\frac \pi 3 (0)\left(-\frac{\sqrt 2} 2\right)\\[6pt] & = -\frac \pi 4 \end{align*}

$$\displaystyle f'(4) = -\frac \pi 4$$

Step 1

Identify the factors in the function.

$$f(x) = \blue{e^{-3x}}\red{\sin 7x}$$

Step 2

Differentiate using the product rule.

\begin{align*} f'(x) & = \blue{-3e^{-3x}}\sin 7x + e^{-3x}\red{(7\cos 7x)}\\[6pt] & = -3e^{-3x}\sin 7x + 7e^{-3x}\cos 7x \end{align*}

Step 3

(Optional) Factor the derivative.

\begin{align*} f'(x) & = -3\blue{e^{-3x}}\sin 7x + 7\blue{e^{-3x}}\cos 7x\\[6pt] & = \blue{e^{-3x}}\left(-3\sin 7x + 7\cos 7x\right)\\[6pt] & = e^{-3x}\left(7\cos 7x-3\sin 7x\right) \end{align*}

$$f'(x) = e^{-3x}\left(7\cos 7x-3\sin 7x\right)$$

Step 1

Identify the factors in the function.

$$f(x) = \blue{e^{-0.5x}}\red{\sin 6x}$$

Step 2

Differentiate using the product rule.

\begin{align*} f'(x) & = \blue{-0.5e^{-0.5x}}\sin 6x + e^{-0.5x}\red{(6\cos 6x)}\\[6pt] & = -0.5e^{-0.5x}\sin 6x + 6e^{-0.5x}\cos 6x \end{align*}

Step 3

(Optional) Factor the derivative.

\begin{align*} f'(x) & = -0.5\blue{e^{-0.5x}}\sin 6x + 6\blue{e^{-0.5x}}\cos 6x\\[6pt] & = \blue{e^{-0.5x}}\left(-0.5\sin 6x + 6\cos 6x\right)\\[6pt] & = 0.5e^{-0.5x}\left(-\sin 6x + 12\cos 6x\right)\\[6pt] & = 0.5e^{-0.5x}\left(12\cos 6x-\sin 6x\right) \end{align*}

Step 4

Evaluate $$f'(0)$$.

\begin{align*} f'(0) & = 0.5e^{-0.5(0)}\left(12\cos 6(0)-\sin 6(0)\right)\\[6pt] & = 0.5e^0\left(12\cos 0-\sin 0\right)\\[6pt] & = 0.5(1)\cdot\left(12(1)-0\right)\\[6pt] & = 0.5(12)\\[6pt] & = 6 \end{align*}

$$f'(0) = 6$$

Step 1

Identify the factors in the function.

$$f(x) = \blue{e^{12x}}\red{\cos\left(\frac \pi 3x\right)}$$

Step 2

Differentiate using the product rule.

\begin{align*} f'(x) & = \blue{12e^{12x}}\cos\left(\frac \pi 3 x\right) + e^{12x}\red{\left(-\frac \pi 3\right)\sin\left(\frac\pi 3 x\right)}\\[6pt] & = 12e^{12x}\cos\left(\frac \pi 3 x\right) -\frac \pi 3 e^{12x}\sin\left(\frac\pi 3 x\right) \end{align*}

Step 3

(Optional) Factor the derivative.

\begin{align*} f'(x) & = 12\blue{e^{12x}}\cos\left(\frac \pi 3 x\right) -\frac \pi 3\blue{e^{12x}}\sin\left(\frac\pi 3 x\right)\\[6pt] & = \blue{e^{12x}}\left[12\cos\left(\frac \pi 3 x\right) -\frac \pi 3\sin\left(\frac\pi 3 x\right)\right]\\[6pt] & = \frac 1 3 e^{12x}\left[36\cos\left(\frac \pi 3 x\right) -\pi\sin\left(\frac\pi 3 x\right)\right] \end{align*}

Step 4

Evaluate the derivative at $$x = 1$$.

\begin{align*} f'(\blue 1) & = \frac 1 3 e^{12\blue{(1)}}\left[36\cos\left(\frac \pi 3 \blue{(1)}\right) -\pi\sin\left(\frac\pi 3 \blue{(1)}\right)\right]\\[6pt] & = \frac 1 3 e^{12}\left[36\cos\left(\frac \pi 3\right) -\pi\sin\left(\frac\pi 3\right)\right]\\[6pt] & = \frac 1 3 e^{12}\left[36\left(\frac 1 2\right) -\pi\left(\frac{\sqrt 3} 2\right)\right]\\[6pt] & = \frac 1 3 e^{12}\left[18 -\frac{\pi\,\sqrt 3} 2\right]\\[6pt] & \approx 828{,}926.5 \end{align*}

$$f'(1) = \frac 1 3 e^{12}\left[18 -\frac{\pi\,\sqrt 3} 2\right] \approx 828{,}926.5$$

Step 1

Identify the factors in the function.

$$f(x) = \blue{e^{-x}}\red{\cos \pi x}$$

Step 2

Differentiate with the product rule.

\begin{align*} f'(x) & = \blue{-e^{-x}}\cos \pi x + e^{-x}\red{(-\pi\sin \pi x)}\\[6pt] & = -e^{-x}\cos \pi x -\pi e^{-x}\sin \pi x \end{align*}

Step 3

(Optional) Factor the derivative.

\begin{align*} f'(x) & = -e^{-x}\cos \pi x -\pi e^{-x}\sin \pi x\\[6pt] & = -e^{-x}\left(\cos \pi x + \pi\sin \pi x\right) \end{align*}

$$f'(x) = -e^{-x}\left(\cos \pi x + \pi\sin \pi x\right)$$

Step 1

Identify the factors in the function.

$$f(x) = \blue{5x^9}\red{e^{-3x}}\sin 2x$$

Step 2

With three factors, we want to remember that "each factor takes turns being the derivative."

\begin{align*} f'(x) & = \big[\blue{5(9x^8)}e^{-3x}\sin 2x\big] + \big[5x^9\blue{(-3e^{-3x})}\sin 2x\big] + \big[5x^9e^{-3x}\blue{(2\cos 2x)}\big]\\[6pt] & = 45x^8e^{-3x}\sin 2x -15x^9e^{-3x}\sin 2x + 10x^9e^{-3x}\cos 2x \end{align*}

Step 3

(Optional) Factor the derivative.

\begin{align*} f'(x) & = 45x^8\blue{e^{-3x}}\sin 2x -15x^9\blue{e^{-3x}}\sin 2x + 10x^9\blue{e^{-3x}}\cos 2x\\[6pt] & = \blue{e^{-3x}}\left(45\red{x^8}\sin 2x -15\red{x^9}\sin 2x + 10\red{x^9}\cos 2x\right)\\[6pt] & = \red{x^8}e^{-3x}\left(45\sin 2x - 15x\sin 2x+ 10x\cos 2x\right)\\[6pt] & = 5x^8e^{-3x}\left(9\sin 2x -3x\sin 2x + 2x\cos 2x\right) \end{align*}

$$f'(x) = 5x^8e^{-3x}\left(9\sin 2x -3x\sin 2x + 2x\cos 2x\right)$$

Step 1

Identify the factors in the function.

$$f(x) = \blue{e^{-0.75x}}\,\red{\sin 4x}\,\cos 8x$$

Step 2

Differentiate using the product rule.

\begin{align*} f'(x) & = \blue{(-0.75e^{-0.75x})}\sin 4x\cos 8x + e^{-0.75x}\blue{(4\cos 4x)}\cos 8x + e^{-0.75x}\sin 4x \blue{(-8\sin 8x)}\\[6pt] & = -0.75e^{-0.75x}\sin 4x\cos 8x + 4e^{-0.75x}\cos 4x\cos 8x -8e^{-0.75x}\sin 4x\sin 8x \end{align*}

Step 3

(Optional) Factor the derivative.

\begin{align*} f'(x) & = -0.75\blue{e^{-0.75x}}\sin 4x\cos 8x + 4\blue{e^{-0.75x}}\cos 4x\cos 8x -8\blue{e^{-0.75x}}\sin 4x\sin 8x\\[6pt] & = \blue{e^{-0.75x}}\left(-0.75\sin 4x\cos 8x + 4\cos 4x\cos 8x -8\sin 4x\sin 8x\right)\\[6pt] & = \frac 1 4 e^{-0.75x}\left(-3\sin 4x\cos 8x + 16\cos 4x\cos 8x -24\sin 4x\sin 8x\right)\\[6pt] & = \frac 1 4 e^{-0.75x}\left(16\cos 4x\cos 8x -24\sin 4x\sin 8x -3\sin 4x\cos 8x\right) \end{align*}

Step 4

Evaluate the derivative at $$x = 0$$.

\begin{align*} f'(\blue 0) & = \frac 1 4 e^{-0.75\blue{(0)}}\left(16\cos 4\blue{(0)}\cos 8\blue{(0)} -24\sin 4\blue{(0)}\sin 8\blue{(0)} -3\sin 4\blue{(0)}\cos 8\blue{(0)}\right)\\[6pt] & = \frac 1 4 e^0\left(16\cos 0\cos 0 -24\sin 0\sin 0 -3\sin 0\cos 0\right)\\[6pt] & = \frac 1 4(1)\left(16(1)(1) -24(0)(0) -3(0)(0)\right)\\[6pt] & = \frac 1 4(16)\\[6pt] & = 4 \end{align*}

$$f'(0) = 4$$

Step 1

Identify the factors in the function.

$$f(x) = \blue{\frac 5 8 x^3}\red{e^x}\cos x$$

Step 2

Differentiate using the product rule.

\begin{align*} f'(x) & = \blue{\frac 5 8\left(3 x^2\right)} e^x\cos x + \frac 5 8 x^3\blue{e^x}\cos x + \frac 5 8 x^3 e^x\blue{(-\sin x)}\\[6pt] & = \frac{15} 8x^2e^x\cos x + \frac 5 8 x^3e^x\cos x - \frac 5 8x^3e^x\sin x \end{align*}

Step 3

(Optional) Factor the derivative.

\begin{align*} f'(x) & = \frac{15} 8x^2\blue{e^x}\cos x + \frac 5 8 x^3\blue{e^x}\cos x - \frac 5 8x^3\blue{e^x}\sin x\\[6pt] & = \blue{e^x}\left(\frac{15} 8\red{x^2}\cos x + \frac 5 8 \red{x^3}\cos x - \frac 5 8\red{x^3}\sin x\right)\\[6pt] & = \red{x^2}e^x\left(\frac{15} 8\cos x + \frac 5 8 x\cos x - \frac 5 8x\sin x\right)\\[6pt] & = \frac 5 8 x^2e^x\left(3\cos x + x\cos x - x\sin x\right) \end{align*}

$$f'(x) = \frac 5 8 x^2 e^x\left(3\cos x + x\cos x - x\sin x\right)$$

Step 1

Identify the factors in the function.

$$f(x) = \blue{3x^2}\red{e^{-10x}}\sin 0.5x$$

Step 2

Differentiate using the product rule.

\begin{align*} f'(x) & = \blue{6x}e^{-10x}\sin 0.5x + 3x^2\blue{(-10e^{-10x})}\sin 0.5x + 3x^2e^{-10x}\blue{(0.5\cos 0.5x)}\\[6pt] & = 6xe^{-10x}\sin 0.5x - 30x^2e^{-10x}\sin 0.5x + 1.5x^2e^{-10x}\cos 0.5x \end{align*}

Step 3

(Optional) Factor the derivative.

\begin{align*} f'(x) & = 6x\blue{e^{-10x}}\sin 0.5x - 30x^2\blue{e^{-10x}}\sin 0.5x + 1.5x^2\blue{e^{-10x}}\cos 0.5x\\[6pt] & = \blue{e^{-10x}}\left(6\red{x}\sin 0.5x - 30\red{x^2}\sin 0.5x + 1.5\red{x^2}\cos 0.5x\right)\\[6pt] & = \red{x}e^{-10x}\left(6\sin 0.5x - 30x\sin 0.5x + 1.5x\cos 0.5x\right)\\[6pt] & = 1.5xe^{-10x}\left(4\sin 0.5x - 20x\sin 0.5x + x\cos 0.5x\right) \end{align*}

$$f'(x) = 1.5xe^{-10x}\left(4\sin 0.5x - 20x\sin 0.5x + x\cos 0.5x\right)$$