2.6 Quotient Rule

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We compute the derivative of a quotient.

The Quotient Rule

If $f(x)$ and $g(x)$ are differentiable then $\left (\frac {f(x)}{g(x)}\right )' = \frac {g(x)f'(x) - f(x)g'(x)}{g^2(x)}$ at all points where $g(x) \neq 0$ .

In words, the derivative of a quotient is the bottom times the derivative of the top minus the top times the derivative of the bottom, all over the bottom squared.

Find $h'(x)$ if $h(x) = \frac {x+1}{x-1}.$ We have $\displaystyle {h(x) = \frac {f(x)}{g(x)}}$ with: $f(x) =x+1 \ \text {and} \ g(x)= x-1.$ To use the quotient rule we need the derivatives: $f'(x) = 1 \ \text {and} \ g'(x) = 1.$ We can now write: $\begin{align*} h'(x) &= \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)}\\ &= \frac{(x-1)(1)- (x+1)(1)}{(x-1)^2}\\ &= \frac{(x-1)-(x+1)}{(x-1)^2} \\ &= -\frac{2}{(x-1)^2}. \end{align*}$

Here is a video of Example 1

Compute $\frac {d}{dx} \left (\frac {x-6}{x+3}\right )$

$\left (\frac {f}{g}\right )' = \frac {f'g-g'f}{g^2}$

The derivative of $x-6$ is $1$

The derivative of $x+3$ is $1$

Collect like terms in the numerator

The derivative of $\frac {x-6}{x+3}$ with respect to $x$ is $\answer [given]{\frac {9}{(x+3)^2}}$

Compute $\frac {d}{dx} \left (\frac {2x+1}{3x-5}\right )$

$\left (\frac {f}{g}\right )' = \frac {f'g-g'f}{g^2}$

The derivative of $2x+1$ is $2$

The derivative of $3x-5$ is $3$

Collect like terms in the numerator

The derivative of $\frac {2x+1}{3x-5}$ with respect to $x$ is $\answer [given]{-\frac {13}{(3x-5)^2}}$

Compute $\frac {d}{dx} \left (\frac {8x-5}{12x - 7}\right )$

$\left (\frac {f}{g}\right )' = \frac {f'g-g'f}{g^2}$

The derivative of $8x-5$ is $8$

The derivative of $12x-7$ is $12$

Collect like terms in the numerator

The derivative of $\frac {8x-5}{12x-7}$ with respect to $x$ is $\answer [given]{\frac {4}{(12x-7)^2}}$

Find $h'(x)$ if $h(x) = \frac {x^2 - 3x}{x^3 +5}.$ We write $\displaystyle {h(x) = \frac {f(x)}{g(x)}}$ with: $f(x) =x^2 - 3x \ \text {and} \ g(x)= x^3 + 5.$ To use the quotient rule we need the derivatives: $f'(x) = 2x-3 \ \text {and} \ g'(x) = 3x^2.$ We can now write: $\begin{align*} h'(x) &= \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)}\\ &= \frac{(x^3 + 5)(2x-3) - (x^2-3x)(3x^2)}{(x^3 + 5)^2}\\ &= \frac{(2x^4 -3x^3 + 10x - 15) -(3x^4-9x^3)}{(x^3 + 5)^2}\\ &= \frac{2x^4 -3x^3 + 10x - 15 -3x^4+9x^3}{(x^3 + 5)^2}\\ &= \frac{-x^4 +6x^3 + 10x - 15}{(x^3 + 5)^2} \end{align*}$

Here is a video of Example 2

Compute $\frac {d}{dx} \left (\frac {x^2 - 2}{x^2 + 1}\right )$

$\left (\frac {f}{g}\right )' = \frac {f'g-g'f}{g^2}$

The derivative of $x^2 - 2$ is $2x$

The derivative of $x^2 + 1$ is $2x$

Collect like terms in the numerator

The derivative of $\frac {x^2 - 2}{x^2 + 1}$ with respect to $x$ is $\answer [given]{\frac {6x}{(x^2 + 1)^2}}$

Compute $\frac {d}{dx} \left (\frac {x^2 + x + 4}{x^2 -3x + 1}\right )$

$\left (\frac {f}{g}\right )' = \frac {f'g-g'f}{g^2}$

The derivative of $x^2 + x + 4$ is $2x+1$

The derivative of $x^2 -3x + 1$ is $2x-3$

Collect like terms in the numerator

The numerator simplifies to $-4x^2 -6x +13$

The derivative of $\frac {x^2 + x + 4}{x^2 -3x + 1}$ with respect to $x$ is $\answer [given]{\frac {-4x^2 -6x +13}{(x^2 -3x + 1)^2}}$

Find $h'(x)$ if $h(x) = \frac {\cos (x)}{2x+1}.$ We write $\displaystyle {h(x) = \frac {f(x)}{g(x)}}$ with: $f(x) = \cos (x) \ \text {and} \ g(x)= 2x+1.$ To use the quotient rule we need the derivatives: $f'(x) = -\sin (x) \ \text {and} \ g'(x) = 2.$ We can now write: $\begin{align*} h'(x) &= \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)}\\ &= \frac{(2x+1)(-\sin(x))- \cos(x)\cdot 2}{(2x+1)^2}\\ &= -\frac{(2x+1)\sin(x) + 2\cos(x)}{(2x+1)^2} \end{align*}$

Here is a video of Example 3

Find $h'(x)$ if $h(x) = \frac {e^x}{x^2 + 1}.$ We write $\displaystyle {h(x) = \frac {f(x)}{g(x)}}$ with: $f(x) =e^x \ \text {and} \ g(x)= x^2 + 1.$ To use the quotient rule we need the derivatives: $f'(x) = e^x \ \text {and} \ g'(x) =2x.$ We can now write: $\begin{align*} h'(x) &= \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)}\\ &= \frac{(x^2+1)e^x - e^x(2x)}{(x^2 + 1)^2} \\ &= \frac{(x^2 - 2x +1)e^x}{(x^2 + 1)^2} \\ &= \frac{e^x(x-1)^2}{(x^2 + 1)^2} \end{align*}$

Here is a video of Example 4

Find $h'(x)$ if $h(x) = \frac {\ln (x)}{x+1}.$ We write $\displaystyle {h(x) = \frac {f(x)}{g(x)}}$ with: $f(x) =\ln (x) \ \text {and} \ g(x)= x+1.$ To use the quotient rule we need the derivatives: $f'(x) =1/x \ \text {and} \ g'(x) = 1.$ We can now write: $\begin{align*} h'(x) &= \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)}\\ &= \frac{(x+1)(1/x) - \ln(x)}{(x+1)^2}\\ &= \frac{(x+1)(1/x) - \ln(x)}{(x+1)^2}\cdot \frac{x}{x}\\ &= \frac{(x+1) - x\ln(x)}{x(x+1)^2}. \end{align*}$

Here is a video of Example 5

Find $h'(x)$ if $h(x) = \tan (x).$ First rewrite $h(x)$ as $\displaystyle {\frac {\sin (x)}{\cos (x)}}.$
Then $\displaystyle {h(x) = \frac {f(x)}{g(x)}}$ with: $f(x) = \sin (x) \ \mbox {and} \ g(x)= \cos (x).$ To use the quotient rule we need the derivatives: $f'(x) = \cos (x) \ \mbox {and} \ g'(x) = -\sin (x).$ We can now write: $\begin{align*} h'(x) &= \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)}\\ &= \frac{\cos(x)\cos(x) - \sin(x)(-\sin(x))}{\cos^2(x)}\\ &= \frac{\cos^2(x)+ \sin^2(x)}{\cos^2(x)}\\ &= \frac{1}{\cos^2(x)} = \sec^2(x) \end{align*}$

This important formula is worth remembering: if $f(x) = \tan (x)$ then $f'(x) =\sec ^2(x)$ .

Here is a video of Example 6

Find $h'(x)$ if $h(x) = \cot (x)$ .
First rewrite $h(x)$ as $\displaystyle {\frac {\cos (x)}{\sin (x)}}$ and then $\displaystyle {h(x) = \frac {f(x)}{g(x)}}$ with: $f(x) = \cos (x) \ \text {and} \ g(x)= \sin (x).$ To use the quotient rule we need the derivatives: $f'(x) = -\sin (x) \ \text {and} \ g'(x) =\cos (x).$ We can now write: $\begin{align*} h'(x) &= \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)}\\ &= \frac{\sin(x)(-\sin(x)) -\cos(x)\cos(x) }{\sin^2(x)}\\ &= -\frac{\sin^2(x)+ \cos^2(x)}{\sin^2(x)}\\ &= -\frac{1}{\sin^2(x)} = -\csc^2(x) \end{align*}$

This important formula is worth remembering: if $f(x) = \cot (x)$ then $f'(x) =-\csc ^2(x)$ .

Here is a video of Example 7

Find $h'(x)$ if $h(x) = \sec (x).$
First we rewrite $h(x)$ as $\displaystyle {\frac {1}{\cos (x)}}$ and then $\displaystyle {h(x) = \frac {f(x)}{g(x)}}$ with: $f(x) = 1 \ \mbox {and} \ g(x)= \cos (x).$ To use the quotient rule we need the derivatives: $f'(x) = 0 \ \mbox {and} \ g'(x) = -\sin (x).$ We can now write: $\begin{align*} h'(x) &= \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)}\\ &= \frac{\cos(x)\cdot 0 - 1\cdot (-\sin(x))}{\cos^2(x)}\\ &= \frac{ \sin(x)}{\cos^2(x)} \\ &= \frac{ 1}{\cos(x)} \cdot \frac{\sin(x)}{\cos(x)}\\ &= \sec(x) \tan(x) \end{align*}$

This important formula is worth remembering: if $f(x) = \sec (x)$ then $f'(x) =\sec (x) \tan (x)$ .

Here is a video of Example 8

Find $h'(x)$ if $h(x) = \frac {x^2 + 3x + 5}{x}.$ We could use the quotient rule, but it is easier to rewrite $h(x)$ as $h(x) = x + 3 + \frac {5}{x} = x + 3 + 5x^{-1}$ and find $h'(x)$ using the power and constant rules. We have: $h'(x) = 1 + (-5)x^{-2} = 1 - \frac {5}{x^2}.$

Here is a detailed, lecture style video on the Quotient Rule: