Vector-Valued Functions and Motion: Background You’ll Need 3

  • Apply specific rules to find derivatives of functions multiplied or divided by each other

The Advanced Rules

The Product Rule

When differentiating a product of two functions, you cannot simply multiply their derivatives. To see why, consider [latex]f(x) = x^2[/latex]. We know [latex]f'(x) = 2x[/latex], but if we incorrectly applied “derivative of product = product of derivatives,” we’d get:

[latex]\dfrac{d}{dx}(x) \cdot \dfrac{d}{dx}(x) = 1 \cdot 1 = 1[/latex]

This is clearly wrong. The correct rule accounts for both functions changing simultaneously.

the product rule

Let [latex]f(x)[/latex] and [latex]g(x)[/latex] be differentiable functions. Then

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

 

That is,

if [latex]j(x)=f(x)g(x)[/latex] then [latex]j^{\prime}(x)=f^{\prime}(x)g(x)+g^{\prime}(x)f(x)[/latex]

 

This means that the derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function.

Remember the Pattern

Some students find it helpful to think: “first derivative, second stays” + “first stays, second derivative”

The order doesn’t matter due to commutativity—you can write the terms in either order:

[latex]f'(x)g(x) + g'(x)f(x) = g'(x)f(x) + f'(x)g(x)[/latex]

For [latex]j(x)=f(x)g(x)[/latex], use the product rule to find [latex]j^{\prime}(2)[/latex] if [latex]f(2)=3, \, f^{\prime}(2)=-4, \, g(2)=1[/latex], and [latex]g^{\prime}(2)=6[/latex].

For [latex]j(x)=(x^2+2)(3x^3-5x)[/latex], find [latex]j^{\prime}(x)[/latex] by applying the product rule. Check the result by first finding the product and then differentiating.

The Quotient Rule

Just as with products, you cannot simply divide derivatives when differentiating a quotient. Consider [latex]f(x) = x^2[/latex]. We know [latex]f'(x) = 2x[/latex], but if we incorrectly applied “derivative of quotient = quotient of derivatives,” we’d get:

[latex]\dfrac{\frac{d}{dx}(x^3)}{\frac{d}{dx}(x)} = \dfrac{3x^2}{1} = 3x^2[/latex]

This is wrong. The quotient rule accounts for the interaction between numerator and denominator.

the quotient rule

Let [latex]f(x)[/latex] and [latex]g(x)[/latex] be differentiable functions. Then

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

 

That is,

if [latex]j(x)=\dfrac{f(x)}{g(x)}[/latex], then [latex]j^{\prime}(x)=\dfrac{f^{\prime}(x)g(x)-g^{\prime}(x)f(x)}{(g(x))^2}[/latex]
Remember: “Low-D-High minus High-D-Low, over Low-Low”

Many students use this mnemonic:

  • Low = denominator function [latex]g(x)[/latex]
  • High = numerator function [latex]f(x)[/latex]
  • D = derivative

So: [latex]\dfrac{\text{(Low)(D-High)} - \text{(High)(D-Low)}}{\text{(Low)}^2}[/latex]

Warning: Order matters here—switching the subtraction will give you the wrong sign.

Use the quotient rule to find the derivative of [latex]k(x)=\dfrac{5x^2}{4x+3}[/latex]

Find the derivative of [latex]h(x)=\dfrac{3x+1}{4x-3}[/latex]

Extended Power Rule

It is now possible to use the quotient rule to extend the power rule to find derivatives of functions of the form [latex]x^k[/latex] where [latex]k[/latex] is a negative integer.

extended power rule

If [latex]k[/latex] is a negative integer, then

[latex]\dfrac{d}{dx}(x^k)=kx^{k-1}[/latex]

Find [latex]\frac{d}{dx}(x^{-4})[/latex]

Use the extended power rule and the constant multiple rule to find [latex]f(x)=\dfrac{6}{x^2}[/latex]