Coverage: Lessons 11-20
Exam Date and Time: Monday, October 19, 8:00pm - 9:00pm
Location: Several, look for Brightspace Announcement
Rules of Differentiation [Lesson 11]
Basic Differentiation Rules
The Derivative of \(e^x\)
Practice Problems
Problem 1. Find \(f'(x)\) for \(f(x) = 5x^4-3x^2+7\).
▼ Show answer\(f'(x)=20x^3-6x\)
Problem 2. Find \(f'(x)\) for \(f(x) = 4e^x+x^3\).
▼ Show answer\(f'(x)=4e^x+3x^2\)
Problem 3. Find the equation of the tangent line to \(f(x)=x^3\) at \(x=2\).
▼ Show answer\(y=12x-16\)
Problem 4. Find \(f'(x)\) for \(f(x) = \dfrac{1}{x^3}\).
▼ Show answer\(f'(x) = -\dfrac{3}{x^4}\)
Problem 5. Find \(f'(x)\) for \(f(x) = \sqrt{x}+e^x\).
▼ Show answer\(f'(x) = \dfrac{1}{2\sqrt{x}}+e^x\)
Product Rule, Quotient Rule & Higher-Order Derivatives [Lesson 12]
Product Rule
Quotient Rule
Higher-Order Derivatives
The second derivative \(f''(x)\) is the derivative of \(f'(x)\); the pattern continues with \(f'''(x)\), \(f^{(4)}(x)\), and in general \(f^{(n)}(x)\).
Practice Problems
Problem 1. Find \(\dfrac{dy}{dx}\) for \(y = \csc(x)\ln(x)\).
▼ Show answer\(\dfrac{dy}{dx} = -\csc(x)\cot(x)\ln(x) + \dfrac{\csc(x)}{x}\)
Problem 2. Find \(f'(x)\) for \(f(x) = \dfrac{x^2+1}{x-1}\).
▼ Show answer\(f'(x) = \dfrac{x^2-2x-1}{(x-1)^2}\)
Problem 3. Find \(f''(x)\) for \(f(x)=x^4-3x^2\).
▼ Show answer\(f''(x)=12x^2-6\)
Problem 4. Find \(f'(x)\) for \(f(x)=(3x-2)(x^2+5)\) using the product rule.
▼ Show answer\(f'(x)=9x^2-4x+15\)
Problem 5. Find \(\dfrac{dy}{dx}\) for \(y = \dfrac{\tan(x)}{x^2+1}\).
▼ Show answer\(\dfrac{dy}{dx} = \dfrac{\sec^2(x)(x^2+1) - 2x\tan(x)}{(x^2+1)^2}\)
Problem 6. Find the 101st derivative of \(f(x)=2xe^x\) at \(x=0\).
▼ Show answer\(f^{(101)}(0)=202\)
Trigonometric Limits & Derivatives [Lesson 13]
Two Key Trig Limits
Derivatives of Trig Functions
| Function | Derivative |
|---|---|
| \(\sin x\) | \(\cos x\) |
| \(\cos x\) | \(-\sin x\) |
| \(\tan x\) | \(\sec^2 x\) |
| \(\cot x\) | \(-\csc^2 x\) |
| \(\sec x\) | \(\sec x\tan x\) |
| \(\csc x\) | \(-\csc x\cot x\) |
Practice Problems
Problem 1. Evaluate \(\displaystyle\lim_{x\to0}\dfrac{\tan(7x)\cos(4x)}{\tan(4x)\cos(7x)}\).
▼ Show answer\(\dfrac{7}{4}\)
Problem 2. Find \(f'(x)\) for \(f(x)=x^2\sin x\).
▼ Show answer\(f'(x)=2x\sin x + x^2\cos x\)
Problem 3. Find \(f'(x)\) for \(f(x)=\dfrac{\tan x}{x}\).
▼ Show answer\(f'(x)=\dfrac{x\sec^2x-\tan x}{x^2}\)
Problem 4. Evaluate \(\displaystyle\lim_{x\to2}\dfrac{\sin(x-2)}{x^2-4}\).
▼ Show answer\(\dfrac14\)
Problem 5. Find \(f'(x)\) for \(f(x)=\sec x\tan x\).
▼ Show answer\(f'(x)=\sec x(\tan^2x+\sec^2x)\)
Problem 6. Treating \(x\) as a constant, evaluate \(\displaystyle\lim_{h\to0}\dfrac{\cos 2x\cos 2h-\sin 2x\sin 2h-\cos 2x}{h}\).
▼ Show answer\(-2\sin 2x\)
Derivatives as Rates of Change [Lesson 14]
Position, Velocity, and Acceleration
If \(s(t)\) is the position of an object at time \(t\), the average velocity on \([a,b]\) is the slope of the secant line through \((a,s(a))\) and \((b,s(b))\), and the instantaneous velocity at \(t=a\) is the slope of the tangent line there.
Practice Problems
Problem 1. An object moves along a line with position \(s(t) = t^3 - 9t^2 + 15t + 25\). Find the velocity function \(v(t)\).
▼ Show answer\(v(t)=3t^2-18t+15\)
Problem 2. Find every time the object is momentarily at rest (\(v(t)=0\)).
▼ Show answer\(t=1\) and \(t=5\)
Problem 3. Find the acceleration function \(a(t)\).
▼ Show answer\(a(t)=6t-18\)
Problem 4. Find when \(a(t)=0\), and describe how the velocity behaves before and after that time.
▼ Show answer\(t=3\); velocity is decreasing for \(t<3\) and increasing for \(t>3\)
Problem 5. Find the initial velocity \(v(0)\) and initial acceleration \(a(0)\).
▼ Show answer\(v(0)=15\), \(a(0)=-18\)
Problem 6. For \(t\ge0\), find the intervals on which the object is moving forward (positive direction) and backward, and the intervals on which the velocity is increasing and decreasing.
▼ Show answerMoving forward on \([0,1)\) and \((5,\infty)\); moving backward on \((1,5)\). Velocity decreasing on \([0,3)\); velocity increasing on \((3,\infty)\)
The Chain Rule [Lessons 14–15]
The Chain Rule
Identify the inside function \(u=g(x)\), differentiate the outside function evaluated at \(u\), then multiply by \(u'\). For a composition of three or more functions, work from the outside in and apply the rule at each layer.
Why it's needed: the Product Rule can differentiate \(\sin^2x\) in one step, but it would take 75 applications to reach \(\sin^{75}x\) that way. Treating \(\sin^{75}x\) as \(u^{75}\) with \(u=\sin x\) and applying the Chain Rule handles it in one step: \(75(\sin x)^{74}\cos x\).
Chaining Multiple Layers
For nested compositions, identify the outermost function first, differentiate it (leaving the inside untouched), then multiply by the derivative of what remains, repeating until every layer has been differentiated.
Practice Problems
Problem 1. Suppose \(f(x)=e^{\sin(7x)}\). Find \(f'(0)\).
▼ Show answer\(f'(0)=7\)
Problem 2. Suppose \(f(x)=\ln[\cos(x^3)]\). Find \(f'(x)\).
▼ Show answer\(f'(x)=-3x^2\tan(x^3)\)
Problem 3. Find \(f'(x)\) for \(f(x)=\tan^4\!\big(\sqrt{x^2+1}\big)\).
▼ Show answer\(f'(x)=4\tan^3\!\big(\sqrt{x^2+1}\big)\sec^2\!\big(\sqrt{x^2+1}\big)\dfrac{x}{\sqrt{x^2+1}}\)
Problem 4. Given the table below and, from a graph, \(g(1)=2\), \(g'(1)=2\), \(g(3)=0\), \(g'(3)=-4\): let \(h(x)=f(g(x))\) and \(k(x)=g(f(x))\). Find \(h'(1)\) and \(k'(3)\).
▼ Show answer| \(x\) | 1 | 2 | 3 |
|---|---|---|---|
| \(f(x)\) | 3 | 2 | 1 |
| \(f'(x)\) | 4 | 5 | 6 |
\(h'(1)=10\); \(k'(3)=12\)
Problem 5. Given \(g(x)=\cos(\pi+f(x))\) with \(f(1)=\dfrac12\) and \(f'(1)=3\), find \(g'(1)\).
▼ Show answer\(g'(1)=3\)
Implicit Differentiation [Lesson 16]
Method
- Differentiate both sides of the equation with respect to \(x\). Treat \(y\) as a function of \(x\), so every term containing \(y\) needs the chain rule: \(\frac{d}{dx}(y^n)=ny^{n-1}y'\), \(\frac{d}{dx}(\sin y)=\cos y\,y'\), and so on. Use the product rule on terms such as \(xy\).
- Collect every term containing \(y'\) on one side of the equation and move all other terms to the other side.
- Factor out \(y'\).
- Divide to solve for \(y'\).
- If a point \((x_0,y_0)\) on the curve is given, substitute it to get the slope, then write the tangent line \(y-y_0=y'(x_0,y_0)\,(x-x_0)\).
Horizontal and vertical tangents: after solving \(y'=\dfrac{N(x,y)}{D(x,y)}\), the tangent is horizontal where \(N=0\) (and \(D\neq0\)) and vertical where \(D=0\) (and \(N\neq0\)). The point must also satisfy the original equation, so substitute the condition \(N=0\) or \(D=0\) back into the curve to find the points.
Practice Problems
Problem 1. Find \(\dfrac{dy}{dx}\) for \(x^2+y^2=25\).
▼ Show answer\(y' = -\dfrac{x}{y}\)
Problem 2. Find \(\dfrac{dy}{dx}\) for \(x^3+y^3=9\).
▼ Show answer\(y' = -\dfrac{x^2}{y^2}\)
Problem 3. Find the equation of the tangent line to the curve \(3y^2+xy^3+x-3=0\) at the point \((0,1)\).
▼ Show answer\(x+3y=3\)
Problem 4. Find the slope of the curve \(2\cos(xy)=y^2-1\) at the point \(\left(\dfrac{\pi}{2},1\right)\).
▼ Show answer\(-\dfrac{2}{2+\pi}\)
Problem 5. Find \(y'\) for the equation \(e^{xy}=3x-y+2\).
▼ Show answer\(y'=\dfrac{3-ye^{xy}}{1+xe^{xy}}\)
Problem 6. Find the equations of all horizontal and vertical tangent lines to the curve \(x^2+xy+y^2=3\).
▼ Show answerHorizontal tangent lines: \(y=2\) and \(y=-2\). Vertical tangent lines: \(x=2\) and \(x=-2\)
Derivatives of Logarithms & Logarithmic Differentiation [Lesson 17]
Basic Log Derivative Rules
Logarithmic Differentiation
Use this method for functions of the form \(y=f(x)^{g(x)}\), and for products or quotients with many factors or large powers.
- Take the natural log of both sides: \(\ln y=\ln\big(f(x)\big)\).
- Expand the right side with log rules: \(\ln(ab)=\ln a+\ln b\), \(\ln\frac ab=\ln a-\ln b\), \(\ln(a^r)=r\ln a\).
- Differentiate both sides with respect to \(x\). The left side becomes \(\dfrac{y'}{y}\) by the chain rule.
- Multiply both sides by \(y\) to solve for \(y'\).
- Replace \(y\) by the original expression in \(x\).
Practice Problems
Problem 1. Find \(f'(x)\) for \(f(x)=\ln(x^2+1)\).
▼ Show answer\(f'(x)=\dfrac{2x}{x^2+1}\)
Problem 2. Find \(f'(x)\) for \(f(x)=x^2\ln x\).
▼ Show answer\(f'(x)=2x\ln x + x\)
Problem 3. Find \(f'(x)\) for \(f(x)=\ln(\sin x)\).
▼ Show answer\(f'(x)=\cot x\)
Problem 4. If \(f(x)=x^{\sin x}\), find \(f'\!\left(\dfrac{\pi}{2}\right)\).
▼ Show answer\(f'\!\left(\tfrac{\pi}{2}\right)=1\)
Problem 5. Use logarithmic differentiation to find \(y'\) for \(y=(x^2+1)^{\sin x}\).
▼ Show answer\(y' = (x^2+1)^{\sin x}\left[\cos x\,\ln(x^2+1) + \dfrac{2x\sin x}{x^2+1}\right]\)
Problem 6. Use logarithmic differentiation to find \(\dfrac{dy}{dx}\) for \(y=x^{x^2}\) (\(x>0\)).
▼ Show answer\(\dfrac{dy}{dx}=x^{x^2+1}(2\ln x+1)\)
Problem 7. Let \(y=\dfrac{(x-1)^3(x+2)^4}{(x+3)^5(x-4)^2}\). Use logarithmic differentiation to find \(y'\), then evaluate \(y'(0)\).
▼ Show answer\(y'=\dfrac{(x-1)^3(x+2)^4}{(x+3)^5(x-4)^2}\left[\dfrac{3}{x-1}+\dfrac{4}{x+2}-\dfrac{5}{x+3}-\dfrac{2}{x-4}\right]\); \(y'(0)=\dfrac{13}{1458}\)
Derivatives of Inverse Trig Functions [Lesson 20]
Key Derivatives
The Inverse Function Derivative Theorem
To use this theorem, first find \(a\) such that \(f(a)=b\) (i.e., \(a=f^{-1}(b)\)), then evaluate \(f'(a)\) and take its reciprocal, without ever needing an explicit formula for \(f^{-1}\).
Practice Problems
Problem 1. Let \(f(x)=\tan^{-1}(x^3+1)\). Find \(f'(1)\).
▼ Show answer\(\dfrac35\)
Problem 2. Find the slope of the line tangent to \(y=x\tan^{-1}x\) at the point \(\left(1,\dfrac{\pi}{4}\right)\).
▼ Show answer\(\dfrac{\pi}{4}+\dfrac12\)
Problem 3. Let \(f(x)=9x+5\). Find \((f^{-1})'(14)\).
▼ Show answer\(\dfrac19\)
Problem 4. Let \(f(x)=8x^3-15x^2-5\). Given \(f(2)=-1\), find \((f^{-1})'(-1)\).
▼ Show answer\(\dfrac{1}{36}\)
Problem 5. Let \(f(x)=x^3+x^2+x+1\). Find \(f^{-1}(4)\) and \((f^{-1})'(4)\).
▼ Show answer\(f^{-1}(4)=1\); \((f^{-1})'(4)=\dfrac16\)
Problem 6. Let \(f(x)=\sin x+x^5+x^3+x+1\). Find \((f^{-1})'(1)\).
▼ Show answer\(\dfrac12\)