NOTE: You may view the original PDFs of the notes written in class by clicking on the link for the 1:30 Section or 2:30 Section, respectively.
Lecture 9: The Quotient Rule; Derivatives of the Other Trigonometric Functions
GOAL: Be able to differentiate functions of the form \( \frac{f(x)}{g(x)} \). Use this to derive the derivatives of all 6 trigonometric functions.
Summary:
\[ \begin{gathered} \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2} \\[0.5em] (\sin x)' = \cos x \qquad (\cos x)' = -\sin x \qquad (\tan x)' = \sec^2 x \\[0.5em] (\sec x)' = \sec x \tan x \qquad (\csc x)' = -\csc x \cot x \qquad (\cot x)' = -\csc^2 x \end{gathered} \]NOTE:
\[ \left[\frac{1}{x}\right]' = \frac{-1}{x^2} \ne 0 = \frac{[1]'}{[x]'} \] So, just as in the Product Rule, the derivative of a quotient is NOT the quotient of the individual derivatives.Q: What's \( \left[\frac{1}{g(x)}\right]' \)?
A: \( \left[\frac{1}{g(x)}\right]' = \frac{-g'(x)}{[g(x)]^2} \) (The derivation using the limit definition is in the textbook)
With this in mind, we can now derive the Quotient Rule
\begin{align*} \left[\frac{f(x)}{g(x)}\right]' =&~ \left[f(x) \cdot \frac{1}{g(x)}\right]'\\[0.5em] =&~ f'(x) \cdot \frac{1}{g(x)} + f(x)\left[\frac{1}{g(x)}\right]'\\[0.5em] =&~ f'(x) \cdot \frac{1}{g(x)} + f(x)\left[\frac{-g'(x)}{[g(x)]^2}\right] \\[0.5em] =&~ \frac{f'(x)}{g(x)} - \frac{f(x) g'(x)}{[g(x)]^2}\\[0.5em] =&~ \frac{f'(x) g(x)}{g(x) g(x)} - \frac{f(x) g'(x)}{[g(x)]^2}\\[0.5em] =&~ \frac{f'(x) g(x) - f(x) g'(x)}{[g(x)]^2} \end{align*}Theorem (Quotient Rule)
Let \( f \) and \( g \) be differentiable functions with \( g(x) \ne 0 \). Then,
\[ \left[\frac{f(x)}{g(x)}\right]' = \frac{f'(x) g(x) - f(x) g'(x)}{[g(x)]^2} \]Here is a way to help memorize the Quotient Rule. If \(f(x)\) is considered the "high" and \(g(x)\) is the "low" function, then the derivative can be expressed as “ low d-high, high d-low, square the low, away we go.”
Example 1
We already know
\[ \left[\frac{1}{x}\right]' = \frac{-1}{x^2} \] Let's verify that, if we use the quotient rule, we get the same answer. \[ \left[\frac{1}{x}\right]' = \frac{[1]' \cdot x - 1 \cdot [x]'}{x^2} = \frac{0 - 1}{x^2} = -\frac{1}{x^2} \]NOTE:
This example highlights a way we can show the power rule is true for \( x^n \) where \( n \) is any integer. We already know it is true if \(n\geq 0\). If it is negative, then:
\[ x^n = \frac{1}{x^{-n}} \]where \( -n \) is a positive integer.
Example 2
Find \( y' \) if \( y = \frac{\sin x}{x^2 + x} \)
\[ \begin{gathered} f(x) = \sin x \longmapsto f'(x) = \cos x \\[0.5em] g(x) = x^2 + x \longmapsto g'(x) = 2x + 1 \\[0.5em] y' = \frac{f'(x) g(x) - f(x) g'(x)}{[g(x)]^2} = \frac{[\cos x][x^2 + x] - [\sin x][2x+1]}{[x^2+x]^2} \end{gathered} \]Example 3
Find \( y' \) if \( y = \frac{2e^t}{e^t - 5} \)
\begin{align*} y' =&~ \frac{[2e^t]'(e^t - 5) - (2e^t)[e^t - 5]'}{(e^t-5)^2} \\[0.5em] =&~ \frac{2e^t(e^t - 5) - 2e^t(e^t)}{(e^t - 5)^2}\\[0.5em] =&~ \frac{2e^{2t} - 10e^t - 2e^{2t}}{(e^t - 5)^2} \\[0.5em] =&~ \frac{-10e^t}{(e^t-5)^2}\\[0.5em] =&~\frac{-10e^t}{e^{2t} - 10e^t + 25} \end{align*}Example 4
Let \( a \) be a constant.
NOTE:
A variable that is treated like a constant is often called a parameter.
You can stop here, but we can simplify this further.
\begin{align*} =&~ \frac{2x^2 - 2ax - x^2 + a^2}{(x-a)^2} \\[0.5em] =&~ \frac{x^2 - 2ax + a^2}{(x-a)^2}\\[0.5em] =&~ \frac{(x-a)^2}{(x-a)^2}\\[0.5em] =&~ 1 \end{align*}Derivatives of Trig Functions
Recall \( [\sin x]' = \cos x \) and \( [\cos x]' = -\sin x \)
Tangent
\[ \begin{gathered} [\tan x]' = \left[\frac{\sin x}{\cos x}\right]' = \frac{[\sin x]'\cos x - \sin x[\cos x]'}{\cos^2 x} \\[0.5em] = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \boxed{\sec^2 x} \end{gathered} \]Example 5
If \( y = \cos x \tan x \), what is \( y' \)?
\[ \begin{gathered} y' = [\cos x]' \tan x + \cos x [\tan x]' \\[0.5em] = -\sin x \tan x + \cos x [\sec^2 x] \end{gathered} \]You can stop here, but we can simplify this further.
\begin{align*} =&~ -\sin x \cdot \frac{\sin x}{\cos x} + \cos x \cdot \frac{1}{\cos^2 x} \\[0.5em] =&~ \frac{-\sin^2 x}{\cos x} + \frac{1}{\cos x}\\[0.5em] =&~ \frac{1 - \sin^2 x}{\cos x} \\[0.5em] =&~ \frac{\cos^2 x}{\cos x}\\[0.5em] =&~ \cos x \end{align*}Here, we needed the fact that \(\sin^2 x + \cos^2 x = 1 \implies \cos^2 x = 1 - \sin^2 x\)
Cotangent
\[ \begin{gathered} [\cot x]' = \left[\frac{\cos x}{\sin x}\right]' = \frac{(-\sin x)\sin x - \cos x(\cos x)}{\sin^2 x} \\[0.5em] = \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} = \frac{-1}{\sin^2 x} = \boxed{-\csc^2 x} \end{gathered} \]Example 6
Find \( y' \) if \( y = e^x \cdot \cot x \)
\[ \begin{gathered} y' = [e^x]' \cot x + e^x [\cot x]' \\[0.5em] = e^x \cot x + e^x(-\csc^2 x) \\[0.5em] = e^x [\cot x - \csc^2 x] \end{gathered} \]Secant
\[ \begin{gathered} [\sec x]' = \left[\frac{1}{\cos x}\right]' = \frac{[1]'\cos x - 1\cdot[\cos x]'}{\cos^2 x} \\[0.5em] = \frac{0 + \sin x}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x}\cdot\frac{\sin x}{\cos x} \\[0.5em] = \boxed{\sec x \tan x} \end{gathered} \]Example 7
What's \( y' \) if \( y = \sec^2 x = [\sec x][\sec x] \)
\[ \begin{gathered} y' = [\sec x]'\sec x + \sec x[\sec x]' \\[0.5em] = (\sec x \tan x)\sec x + \sec x(\sec x \tan x) \\[0.5em] = 2\sec x (\sec x \tan x) \end{gathered} \]Note that the final answer can be written in a particular way, i.e., \( \frac{d}{d(\sec x)}\left[(\sec x)^2\right] \) \( \frac{d}{dx}(\sec x) \). So, this at least suggests there is a way to compute the derivative of \(\sec^2 x\) without having to break it up into a product.
Cosecant
\[ \begin{gathered} [\csc x]' = \left[\frac{1}{\sin x}\right]' = \frac{0\cdot\sin x - 1\cdot\cos x}{\sin^2 x} \\[0.5em] = -\frac{\cos x}{\sin^2 x} = -\frac{1}{\sin x}\cdot\frac{\cos x}{\sin x} = \boxed{-\csc x \cot x} \end{gathered} \]Example 8
Find \( y' \) if \( y = x^5 \csc x \)
\[ \begin{gathered} y' = 5x^4 \csc x + x^5 [\csc x]' \\[0.5em] = 5x^4 \csc x + x^5 [-\csc x \cot x] \\[0.5em] = x^4 \csc x [5 - x \cot x] \end{gathered} \]Example 9
Find \( h'(7) \) if \( h(x) = \frac{x^2 - 14}{10 - x} \)
\( f(x) = x^2 - 14 \)
\( g(x) = 10 - x \)
\( f(7) = 49 - 14 = 35 \)
\( f'(x) = 2x \)
\( f'(7) = 14 \)
\( g(7) = 10 - 7 = 3 \)
\( g'(x) = -1 \)
\( g'(7) = -1 \)
\( [g(7)]^2 = 9 \)
\[ \begin{gathered} h'(7) = \frac{f'(7)g(7) - f(7)g'(7)}{[g(7)]^2} = \frac{14 \cdot 3 - 35 \cdot (-1)}{9} \\[0.5em] = \frac{42 + 35}{9} = \frac{77}{9} \end{gathered} \]