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.

example 1 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

(problem 1a) 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 $\answer {\frac {9}{(x+3)^2}}$

(problem 1b) 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 $\answer {-\frac {13}{(3x-5)^2}}$

(problem 1c) 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 $\answer {\frac {4}{(12x-7)^2}}$

example 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

(problem 2a) 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 $\answer {\frac {6x}{(x^2 + 1)^2}}$

(problem 2b) 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 $\answer {\frac {-4x^2 -6x +13}{(x^2 -3x + 1)^2}}$

example 3 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

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

The derivative of $\frac {\sin (x)}{x^2 + 1}$ with respect to $x$ is $\answer {\frac {(x^2 + 1)\cos (x) - 2x\sin (x)}{(x^2 + 1)^2}}$

example 4 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

(problem 4) $\displaystyle {\frac {d}{dx} \left (\frac {3e^x}{x + 1}\right )=}$

$\displaystyle {\frac {3xe^x+6e^x}{(x+1)^2}}$ $\displaystyle {\frac {3xe^x}{(x+1)}}$ $\displaystyle {\frac {3xe^x}{(x+1)^2}}$

example 5 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

(problem 5) $\displaystyle {\frac {d}{dx} \left (\frac {2\ln (x)}{3x + 2}\right )=}$

$\displaystyle {\frac {2 - 6\ln (x)}{x(3x+2)}}$ $\displaystyle {\frac {6x+4 - 6\ln (x)}{x(3x+2)^2}}$ $\displaystyle {\frac {6x+4 - 6x\ln (x)}{x(3x+2)^2}}$

example 6 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

(problem 6a) $\displaystyle {\frac {d}{dx} 5\tan (x)= \answer {5\sec ^2(x)}}$

(problem 6b) $\displaystyle {\frac {d}{dx} x\tan (x)= \answer {\tan (x) + x\sec ^2(x)}}$

(problem 6c) Find the equation of the tangent line to $y = \tan (x)$ at $x = 0$ .
The equation is $y = \answer {x}$

example 7 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

(problem 7a) $\displaystyle {\frac {d}{dx} 3\cot (x)= \answer {-3\csc ^2(x)}}$

(problem 7b) $\displaystyle {\frac {d}{dx} x^3\cot (x)= \answer {3x^2\cot (x) - x^3\csc ^2(x)}}$

example 8 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

(problem 8a) $\displaystyle {\frac {d}{dx} 4\sec (x)= \answer {4\sec (x)\tan (x)}}$

(problem 8b) $\displaystyle {\frac {d}{dx} 2e^x \sec (x)= \answer {2e^x\sec (x)(1+\tan (x))}}$

(problem 8c) Find the equation of the tangent line to $y = \sec (x)$ at $x = 0$ .
The equation is $y = \answer {1}$

example 9 Find $h'(x)$ if $h(x) = \csc (x).$
First we rewrite $h(x)$ as $\displaystyle {\frac {1}{\sin (x)}}$ and then $\displaystyle {h(x) = \frac {f(x)}{g(x)}}$ with: $f(x) = 1 \ \mbox {and} \ g(x)= \sin (x).$ To use the quotient rule we need the derivatives: $f'(x) = 0 \ \mbox {and} \ g'(x) = \cos (x).$ We can now write: $\begin{align*} h'(x) &= \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)}\\ &= \frac{\sin(x)\cdot 0 - 1\cdot (\cos(x))}{\sin^2(x)}\\ &= \frac{ -\cos(x)}{\sin^2(x)} \\ &= -\frac{ 1}{\sin(x)} \cdot \frac{\cos(x)}{\sin(x)}\\ &= -\csc(x) \cot(x) \end{align*}$

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

(problem 9a) $\displaystyle {\frac {d}{dx} \pi \csc (x)= \answer {-\pi \csc (x)\cot (x)}}$

(problem 9b) $\displaystyle {\frac {d}{dx} \ln (x) \csc (x)= \answer {\csc (x)/x - \ln (x)\csc (x)\cot (x)}}$

(problem 9c) Find the equation of the tangent line to $y = \csc (x)$ at $x = \pi /2$ .
The equation is $y = \answer {1}$

example 10 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: $h'(x) = 1 + (-5)x^{-2} = 1 - \frac {5}{x^2}.$

(problem 10a) Find $h'(x)$ if $h(x) = \frac {x^3 - 4x + 3}{x^2}.$ Rewrite $h(x)$ as

$x - 4x+ 3$ $x - 4x^{-1} + 3x^{-2}$ $x^3 - 4x + \frac {3}{x^2}$

$h'(x) = \answer {1 + 4x^{-2} -6x^{-3}}$

(problem 10b) Find $h'(x)$ if $h(x) = \frac {x - 2\sqrt {x} + 6}{2x}.$ Rewrite $h(x)$ as

$\frac 12 - x^{-1/2} + 3 x^{-1}$ $\frac 12 - 2\sqrt {x} + 6$ $\frac 12 - x^{-3/2} + 3 x^{-1}$

$h'(x) = \answer {\frac 12 x^{-3/2} -3x^{-2}}$

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