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Wize University Calculus 1 Textbook > Derivatives

The Product Rule

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The Product Rule

Our next differentiation shortcut lets us take derivatives of the product (multiplication) of two functions.

Product Rule

The derivative of a product y=f(x)g(x)y=f\left(x\right)g\left(x\right), where f(x)f(x) and g(x)g(x) are differentiable functions, is:

[f(x)g(x)]′=f′(x)g(x)+f(x)g′(x)\boxed{\quad \left[f\left(x\right)g\left(x\right)\right]'=f'\left(x\right)g\left(x\right)+f\left(x\right)g'\left(x\right)\quad }

Wize Tip
A nice way to remember the product rule is: "the derivative of the 1st times the 2nd, plus the 2nd times the derivative of the 1st".

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Example: The Product Rule

Find the following derivative

ddx[x2(x+7)]\displaystyle\mydd{x^2(x+7)}{x}[bt]


ddx[x2(x+7)]=ddx[x2](x+7)+(x2)ddx[x+7]=(2x)(x+7)+x2(1)=2x2+14x+x2=3x2+14x\displaystyle\mydd{x^2(x+7)}{x}[bt] = \mydd{x^2}{x}[bt](x+7) + (x^2) \mydd{x+7}{x}[bt] \\ \text{} \\ = (2x)(x+7) + x^2(1)=2x^2+14x+x^2 \\ \text{} \\ =\boxed{3x^2+14x}




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Example: The Product Rule

Find the following derivative using the product rule

ddx(x2+1)x2\displaystyle\frac{d}{dx}\frac{(x^2+1)}{x^2}


ddx(x2+1)x2\displaystyle\frac{d}{dx}\frac{(x^2+1)}{x^2} =ddx(x2+1)x−2=\displaystyle\frac{d}{dx}(x^2+1)x^{-2}
=(x2+1)′×x−2+(x2+1)×(x−2)′\displaystyle=(x^2+1)'\times x^{-2}+(x^2+1)\times (x^{-2})'
=2x×x−2+(x2+1)(−2x−3)\displaystyle=2x\times x^{-2}+(x^2+1)(-2x^{-3})
=2x−2(x2+1)x3=\displaystyle\frac{2}{x}-\frac{2(x^2+1)}{x^3}
Find the derivative of y=xf(x)y=\displaystyle xf(x) at x=1x=1 if f(1)=2f(1)=2 and f′(1)=3f'(1)=3


Extra Practice
Product Rule
Suppose g(x)g\left(x\right) is differentiable, f(x)=x2g(x)f\left(x\right)=x^2g\left(x\right). Given that f′(3)=30f'\left(3\right)=30, f′′(3)=19f''\left(3\right)=19, g(3)=2g\left(3\right)=2, what is g′(3)g'\left(3\right) and g′′(3)g''\left(3\right)?
Product Rule
Suppose g(x)g\left(x\right) is differentiable, f(x)=x2g(x)f\left(x\right)=x^2g\left(x\right). Given that f′(3)=30f'\left(3\right)=30, f′′(3)=19f''\left(3\right)=19, g(3)=2g\left(3\right)=2, what is g′(3)g'\left(3\right) and g′′(3)g''\left(3\right)?
Product Rule
Suppose g(x)g\left(x\right) is differentiable, f(x)=x2g(x)f\left(x\right)=x^2g\left(x\right). Given that f′(3)=30f'\left(3\right)=30, f′′(3)=19f''\left(3\right)=19, g(3)=2g\left(3\right)=2, what is g′(3)g'\left(3\right) and g′′(3)g''\left(3\right)?
Product Rule
Suppose g(x)g\left(x\right) is differentiable, f(x)=x2g(x)f\left(x\right)=x^2g\left(x\right). Given that f′(3)=30f'\left(3\right)=30, f′′(3)=19f''\left(3\right)=19, g(3)=2g\left(3\right)=2, what is g′(3)g'\left(3\right) and g′′(3)g''\left(3\right)?
Product Rule
Suppose g(x)g\left(x\right) is differentiable, f(x)=x2g(x)f\left(x\right)=x^2g\left(x\right). Given that f′(3)=30f'\left(3\right)=30, f′′(3)=19f''\left(3\right)=19, g(3)=2g\left(3\right)=2, what is g′(3)g'\left(3\right) and g′′(3)g''\left(3\right)?
Product Rule
Suppose g(x)g\left(x\right) is differentiable, f(x)=x2g(x)f\left(x\right)=x^2g\left(x\right). Given that f′(3)=30f'\left(3\right)=30, f′′(3)=19f''\left(3\right)=19, g(3)=2g\left(3\right)=2, what is g′(3)g'\left(3\right) and g′′(3)g''\left(3\right)?
Product Rule
Suppose g(x)g\left(x\right) is differentiable, f(x)=x2g(x)f\left(x\right)=x^2g\left(x\right). Given that f′(3)=30f'\left(3\right)=30, f′′(3)=19f''\left(3\right)=19, g(3)=2g\left(3\right)=2, what is g′(3)g'\left(3\right) and g′′(3)g''\left(3\right)?
Product Rule
Suppose g(x)g\left(x\right) is differentiable, f(x)=x2g(x)f\left(x\right)=x^2g\left(x\right). Given that f′(3)=30f'\left(3\right)=30, f′′(3)=19f''\left(3\right)=19, g(3)=2g\left(3\right)=2, what is g′(3)g'\left(3\right) and g′′(3)g''\left(3\right)?
Product Rule
Suppose g(x)g\left(x\right) is differentiable, f(x)=x2g(x)f\left(x\right)=x^2g\left(x\right). Given that f′(3)=30f'\left(3\right)=30, f′′(3)=19f''\left(3\right)=19, g(3)=2g\left(3\right)=2, what is g′(3)g'\left(3\right) and g′′(3)g''\left(3\right)?
Product Rule
Suppose g(x)g\left(x\right) is differentiable, f(x)=x2g(x)f\left(x\right)=x^2g\left(x\right). Given that f′(3)=30f'\left(3\right)=30, f′′(3)=19f''\left(3\right)=19, g(3)=2g\left(3\right)=2, what is g′(3)g'\left(3\right) and g′′(3)g''\left(3\right)?
Product Rule
Suppose g(x)g\left(x\right) is differentiable, f(x)=x2g(x)f\left(x\right)=x^2g\left(x\right). Given that f′(3)=30f'\left(3\right)=30, f′′(3)=19f''\left(3\right)=19, g(3)=2g\left(3\right)=2, what is g′(3)g'\left(3\right) and g′′(3)g''\left(3\right)?
Product Rule
Suppose g(x)g\left(x\right) is differentiable, f(x)=x2g(x)f\left(x\right)=x^2g\left(x\right). Given that f′(3)=30f'\left(3\right)=30, f′′(3)=19f''\left(3\right)=19, g(3)=2g\left(3\right)=2, what is g′(3)g'\left(3\right) and g′′(3)g''\left(3\right)?
Product Rule
Suppose g(x)g\left(x\right) is differentiable, f(x)=x2g(x)f\left(x\right)=x^2g\left(x\right). Given that f′(3)=30f'\left(3\right)=30, f′′(3)=19f''\left(3\right)=19, g(3)=2g\left(3\right)=2, what is g′(3)g'\left(3\right) and g′′(3)g''\left(3\right)?
Product Rule
Suppose g(x)g\left(x\right) is differentiable, f(x)=x2g(x)f\left(x\right)=x^2g\left(x\right). Given that f′(3)=30f'\left(3\right)=30, f′′(3)=19f''\left(3\right)=19, g(3)=2g\left(3\right)=2, what is g′(3)g'\left(3\right) and g′′(3)g''\left(3\right)?
Product Rule
Suppose g(x)g\left(x\right) is differentiable, f(x)=x2g(x)f\left(x\right)=x^2g\left(x\right). Given that f′(3)=30f'\left(3\right)=30, f′′(3)=19f''\left(3\right)=19, g(3)=2g\left(3\right)=2, what is g′(3)g'\left(3\right) and g′′(3)g''\left(3\right)?
Product Rule
Q:\textbf{Q:} Differentiate f(x)=(x3/2)(2x4−5x2−x−1)(xe−π3.5)\displaystyle f(x)=(x^{3/2})\left(2x^{4}-5x^{2}-x^{-1}\right)(x^{e}-\pi^{3.5})
Quotient and Product
Find the derivative of g(t)=(1+2t7t)(t2−1)\displaystyle g(t)=\left(\frac{1+2t}{7t}\right)(t^{2}-1)
The product rule
Find the derivative of f(x)=x3exf(x) = x^3 e^x
The graph of w(x) is given below.
Practice: Product Rule*
Suppose g(x)g\left(x\right) is differentiable, f(x)=x2g(x)f\left(x\right)=x^2g\left(x\right). Given that f′(3)=30f'\left(3\right)=30, f′′(3)=19f''\left(3\right)=19, g(3)=2g\left(3\right)=2, what is g′(3)g'\left(3\right) and g′′(3)g''\left(3\right)?
Product Rule
Differentiate f(x)=(x3/2)(2x4−5x2−x−1)(xe−π3.5)\displaystyle f(x)=(x^{3/2})\left(2x^{4}-5x^{2}-x^{-1}\right)(x^{e}-\pi^{3.5})
Quotient and Product
Find the derivative of g(t)=(1+2t7t)(t2−1)\displaystyle g(t)=\left(\frac{1+2t}{7t}\right)(t^{2}-1)
Find the derivative of f(x)=(x3/2+2x)(x7/4+x)\displaystyle f(x)=\left(x^{3/2}+\frac{2}{x}\right)\left(x^{7/4}+x\right) .
Product Rule
Suppose g(x)g\left(x\right) is differentiable, f(x)=x2g(x)f\left(x\right)=x^2g\left(x\right). Given that f′(3)=30f'\left(3\right)=30, f′′(3)=19f''\left(3\right)=19, g(3)=2g\left(3\right)=2, what is g′(3)g'\left(3\right) and g′′(3)g''\left(3\right)?