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Use the product rule to find the derivative of the following functions.

\(\textbf{1)}\) \(f(x)=5x^{3} \cos x\)

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The derivative is \(f'(x)=-5x^{3} \sin x+15x^{2} \cos x\)

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\(\,\,\,\,\,\,f(x)=5x^{3} \cos x\)

\(\,\,\,\,\,\,\frac{d}{dx}[f(x)⋅g(x)]=f(x) \cdot g′(x)+f′(x) \cdot g(x)\)

\(\,\,\,\,\,\,f'(x)=5\left[\left(x^{3}\right) \cdot \left(\cos x\right)’+\left(x^{3}\right)’ \cdot \left(\cos x\right)\right]\)

\(\,\,\,\,\,\,f'(x)=5\left[\left(x^{3}\right) \cdot \left(-\sin x\right)+\left(3x^{2}\right) \cdot \left(\cos x\right)\right]\)

\(\,\,\,\,\,\,\)The derivative is \(f'(x)=-5x^{3} \sin x+15x^{2} \cos x\)

\(\textbf{2)}\) \(f(x)=\sqrt{x} (x^{2}+5)\)

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The derivative is \(f'(x)=\displaystyle\frac{5x^{2}+5}{2\sqrt{x}}\)

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\(\,\,\,\,\,\,f(x)=\sqrt{x} (x^{2}+5)\)

\(\,\,\,\,\,\,\frac{d}{dx}[f(x)⋅g(x)]=f(x) \cdot g′(x)+f′(x) \cdot g(x)\)

\(\,\,\,\,\,\,f'(x)=\left(\sqrt{x}\right)\left(x^{2}+5\right)’+\left(\sqrt{x}\right)’\left(x^{2}+5\right)\)

\(\,\,\,\,\,\,f'(x)=\left(\sqrt{x}\right)\left(2x\right)+\left(\displaystyle\frac{1}{2\sqrt{x}}\right)\left(x^{2}+5\right)\)

\(\,\,\,\,\,\,\displaystyle f'(x)=\frac{2x\sqrt{x}}{1}+\frac{x^{2}+5}{2\sqrt{x}}\)

\(\,\,\,\,\,\,\displaystyle f'(x)=\frac{4x^2}{2\sqrt{x}}+\frac{x^{2}+5}{2\sqrt{x}}\)

\(\,\,\,\,\,\,\displaystyle f'(x)=\frac{4x^2 + x^{2}+5}{2\sqrt{x}}\)

\(\,\,\,\,\,\,\displaystyle f'(x)=\frac{5x^{2}+5}{2\sqrt{x}}\)

\(\textbf{3)}\) \(f(x)=(x^{3}+2)(x^{2}-1)\)

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The derivative is \(f'(x)=5x^{4}-3x^{2}+4x\)

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Let’s find the derivative using the product rule:

\(\,\,\,\,\,\,\frac{d}{dx}[f(x)⋅g(x)]=f(x) \cdot g′(x)+f′(x) \cdot g(x)\)

\(\,\,\,\,\,\,f'(x)=(x^{3}+2)(x^{2}-1)’+(x^{3}+2)'(x^{2}-1)\)

\(\,\,\,\,\,\,f'(x)=(x^{3}+2)(2x)+(3x^{2})(x^{2}-1)\)

\(\,\,\,\,\,\,f'(x)=2x^{4}+4x+3x^{4}-3x^{2}\)

\(\,\,\,\,\,\,f'(x)=5x^{4}-3x^{2}+4x\)

\(\textbf{4)}\) \(f(x)=(x^{4}+4x)(x^{2}-2)\)

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The derivative is \(f'(x)=6x^{5}-8x^{3}+12x^{2}-8\)

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\(\,\,\,\,\,\,\frac{d}{dx}[f(x)⋅g(x)]=f(x) \cdot g′(x)+f′(x) \cdot g(x)\)

\(\,\,\,\,\,\,f'(x)=(x^{4}+4x)(x^{2}-2)’+(x^{4}+4x)'(x^{2}-2)\)

\(\,\,\,\,\,\,f'(x)=(x^{4}+4x)(2x)+(4x^{3}+4)(x^{2}-2)\)

\(\,\,\,\,\,\,f'(x)=2x^{5}+8x^{2}+4x^{5}-8x^{3}+4x^{2}-8)\)

\(\,\,\,\,\,\,f'(x)=6x^{5}-8x^{3}+12x^{2}-8\)

\(\textbf{5)}\) \(f(x)=3e^{x}\sqrt{x}\)

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The derivative is \(f'(x)=\displaystyle\frac{3e^{x}}{2\sqrt{x}}+3e^{x}\sqrt{x}\)

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\(\,\,\,\,\,\,f(x)=3e^{x}\sqrt{x}\)

\(\,\,\,\,\,\,\frac{d}{dx}\left[f(x)⋅g(x)\right]=f(x) \cdot g′(x)+f′(x) \cdot g(x)\)

\(\,\,\,\,\,\,f'(x)=\left(3e^{x}\right)\left(\sqrt{x}\right)’+\left(3e^{x}\right)’\left(\sqrt{x}\right)\)

\(\,\,\,\,\,\,f'(x)=\left(3e^{x}\right)\left(\frac{1}{2\sqrt{x}}\right)+\left(3e^{x}\right)\left(\sqrt{x}\right)\)

\(\,\,\,\,\,\,f'(x)=\frac{3e^{x}}{2\sqrt{x}}+3e^{x}\sqrt{x}\)

\(\textbf{6)}\) Find \(f'(3)\) where \(f(x)=e^{x} g(x),\) \(\,g(3)=6\) and \(\,g'(3)=2\)

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The answer is \(f'(3)=8e^{3}\)

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\(\,\,\,\,\,\,f(x)=e^{x} \cdot g(x)\)

\(\,\,\,\text{Step 1: Take the derivative using Product Rule}\)

\(\,\,\,\,\,\,\frac{d}{dx}[f(x)⋅g(x)]=f(x) \cdot g′(x)+f′(x) \cdot g(x)\)

\(\,\,\,\,\,\,f'(x)=e^{x} \cdot g′(x)+\left(e^{x}\right)’ \cdot g(x)\)

\(\,\,\,\,\,\,f'(x)=e^{x} \cdot g′(x)+e^{x} \cdot g(x)\)

\(\,\,\,\text{Step 2: Plug in }x=3\)

\(\,\,\,\,\,\,f'(3)=e^{3} \cdot g′(3)+e^{3} \cdot g(3)\)

\(\,\,\,\,\,\,f'(3)=e^{3} \cdot (2) + e^{3} \cdot (6)\)

\(\,\,\,\,\,\,f'(3)=2e^{3} + 6e^{3}\)

\(\,\,\,\,\,\,f'(3)=8e^{3}\)

\(\,\,\,\,\,\,\)The answer is \(f'(3)=8e^{3}\)

\(\textbf{7)}\) \(f(x)=x \ln⁡x\)

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The answer is \(f'(x)=\ln⁡x+1\)

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Let’s find the derivative using the product rule:

\(\,\,\,\,\,\,\frac{d}{dx}[f(x)⋅g(x)]=f(x) \cdot g′(x)+f′(x) \cdot g(x)\)

\(\,\,\,\,\,\,f'(x)=x \cdot (\ln⁡x)’+x’\cdot \ln⁡x\)

\(\,\,\,\,\,\,f'(x)=x \cdot \left(\frac{1}{x}\right)+1 \cdot \ln⁡x\)

\(\,\,\,\,\,\,f'(x)=\ln⁡x+1\)

\(\textbf{8)}\) \(f(x)=(2x^{3}+3)(x^{2}-2x+1)\)

Show Answer

The derivative is \(f'(x)=10x^{4}-16x^{3}+6x^{2}+6x-6\)

Show Work

Let’s find the derivative using the product rule:

\(\,\,\,\,\,\,\frac{d}{dx}[f(x)⋅g(x)]=f(x) \cdot g′(x)+f′(x) \cdot g(x)\)

\(\,\,\,\,\,\,f'(x)=(2x^{3}+3)(x^{2}-2x+1)’+(2x^{3}+3)'(x^{2}-2x+1)\)

\(\,\,\,\,\,\,f'(x)=(2x^{3}+3)(2x-2)+(6x^{2})(x^{2}-2x+1)\)

\(\,\,\,\,\,\,f'(x)=4x^{4}-4x^{3}+6x-6+6x^{4}-12x^{3}+6x^{2}\)

\(\,\,\,\,\,\,f'(x)=10x^{4}-16x^{3}+6x^{2}+6x-6\)

See Related Pages\(\)

\(\bullet\text{ Calculus Homepage}\)
\(\,\,\,\,\,\,\,\,\text{All the Best Topics…}\)

\(\bullet\text{ Definition of Derivative}\)
\(\,\,\,\,\,\,\,\, \displaystyle \lim_{\Delta x\to 0} \frac{f(x+ \Delta x)-f(x)}{\Delta x} \)

\(\bullet\text{ Equation of the Tangent Line}\)
\(\,\,\,\,\,\,\,\,f(x)=x^3+3x^2−x \text{ at the point } (2,18)\)

\(\bullet\text{ Derivatives- Constant Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(c)=0\)

\(\bullet\text{ Derivatives- Power Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(x^n)=nx^{n-1}\)

\(\bullet\text{ Derivatives- Constant Multiple Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(cf(x))=cf'(x)\)

\(\bullet\text{ Derivatives- Sum and Difference Rules}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(x) \pm g(x)]=f'(x) \pm g'(x)\)

\(\bullet\text{ Derivatives- Sin and Cos}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}sin(x)=cos(x)\)

\(\bullet\text{ Derivatives- Product Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(x) \cdot g(x)]=f(x) \cdot g'(x)+f'(x) \cdot g(x)\)

\(\bullet\text{ Derivatives- Quotient Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\displaystyle\frac{f(x)}{g(x)}\right]=\displaystyle\frac{g(x) \cdot f'(x)-f(x) \cdot g'(x)}{[g(x)]^2}\)

\(\bullet\text{ Derivatives- Chain Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(g(x))]= f'(g(x)) \cdot g'(x)\)

\(\bullet\text{ Derivatives- ln(x)}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[ln(x)]= \displaystyle \frac{1}{x}\)

\(\bullet\text{ Implicit Differentiation}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Horizontal Tangent Line}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Mean Value Theorem}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Related Rates}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Increasing and Decreasing Intervals}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Intervals of concave up and down}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Inflection Points}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Graph of f(x), f'(x) and f”(x)}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Newton’s Method}\)
\(\,\,\,\,\,\,\,\,x_{n+1}=x_n – \displaystyle \frac{f(x_n)}{f'(x_n)}\)

In Summary

The product rule is a fundamental principle in calculus that allows us to differentiate functions that are the product of two or more functions. It states that the derivative of the product of two functions is equal to the first function times the derivative of the second function plus the second function times the derivative of the first function.

The product rule is an essential tool for finding the derivative of complex functions and is used in a wide range of applications, including engineering, physics, and economics. It is a fundamental building block of calculus and is taught in calc 1 along with other derivative methods. It is a fundamental principle that is used throughout the study of calculus and is essential for understanding more advanced concepts and techniques.

One common mistake when using the product rule is forgetting to apply the chain rule when one of the factors is itself a composite function. It is important to carefully consider the structure of the function being differentiated and apply the appropriate rules to each component.

The product rule is closely related to several other principles in calculus, including the quotient rule, which is used to differentiate functions that are the ratio of two other functions, and the chain rule, which is used to differentiate composite functions.

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