Calculus Cheatsheet
1. Derivatives of Polynomials
- Given a polynomial term of the form \(ax^b\) (where a is a constant coefficient and \(b\) is a constant exponent), to take a derivative with respect to the variable \(x\), first multiply the coefficient \(a\) by the exponent \(b\), and then reduce the exponent by \(1\) according to the following formula:
\[ \dfrac{d}{dx}(ax^b) = bax^{b-1} \]
Note that a few special exponents can make a polynomial look somewhat different: 2. If you see a variable in a denominator, you may bring it to the numerator by negating its exponent according to the following rule:
\[ \dfrac{1}{x^n}= x^{-n} \]
- If you see a square root, rewrite it using an exponent of \(\dfrac 1 2\)
\[ \sqrt{x} = x^{\frac 1 2} \]
- If no coefficient or exponent is observed, the number \(1\) is implied:
\[ x = 1x^1 \]
\[ \dfrac 1 x = \dfrac{1}{x^1} = x^{-1} \]
- It may help to recall the following rules of algebra regarding exponents:
\[ \begin{align*} x^mx^n &= x^{m+n}\\ \dfrac{x^m}{x^n} &= x^{m-n}\\ x^0 &=1\\ (x^m)^n &= x^{mn}\\ (ax)^m&= a^mx^m\\ (a^mx^n)^p &= a^{mp}x^{np}\\ \sqrt{ax} &= (ax)^{\frac 1 2} = a^{\frac 1 2}x^{\frac 1 2} \end{align*} \]
- If a polynomial includes multiple terms, find the derivative of each term individually and then add the results together:
\[ \dfrac{d}{dx}(y_1 + y_2 + \dots + y_n) = \dfrac{dy_1}{dx}+\dfrac{dy_2}{dx}+ \cdots+\dfrac{dy_n}{dx} \]
- the derivative of a constant equals zero:
\[ \dfrac{d}{dx}c=0 \]
2. Chain rule, Product rule, Quotient rule
- If \(f(u)\) is a function of one variable \(u\), and if \(u(x)\) is itself a function of a second variable \(x\), then the chain rule may be applied in order to find a derivative of the function \(f(u(x))\) with respect to the second variable \(x\):
\[ \dfrac{df}{dx} = \dfrac{df}{du} \dfrac{du}{dx} \]
- Note the following:
- f(u) is called the outside function.
- u(x) is called the inside function.
- The chain rule basically states that df/dx equals the product of the derivatives of the outside function (df/du) and the inside function (du/dx).
- The chain rule is often helpful when taking derivatives of functions like these:
\[ \begin{align*} & \dfrac{d}{dx} (3x-2)^{20}\\ & \dfrac{d}{dx}\sqrt{2x^2+1}\\ & \dfrac{d}{dx} \dfrac{1}{x^2-4x+5}\\ &\dfrac{d}{dx}\sin(4x^3+3) \end{align*} \]
- The product rule allows you to take a derivative of two different functions, \(f(x)\) and \(g(x)\), of the same argument \((x)\), when the functions are multiplied together:
\[ \dfrac{d}{dx}(fg) = g\dfrac{df}{dx}+f\dfrac{dg}{dx} \]
- The quotient rule allows you to take a derivative of two different functions, \(f(x)\) and \(g(x)\), of the same argument \((x)\), when the functions are being divided:
\[ \dfrac{d}{dx}\left(\dfrac f g\right) = \dfrac{g \dfrac{df}{dx}-f\dfrac{dg}{dx}}{g^2} \]
- The product rule is useful when taking derivatives like these:
\[ \begin{align*} &\dfrac{d}{dx}(2x+3)^8(x^2-1)^5\\ &\dfrac{d}{dx}(5x^2-2)\sqrt{8x+3}\\ &\dfrac{d}{dx}x^3 \sin(2x) \end{align*} \]
- The quotient rule is useful when taking derivatives like these:
\[ \begin{align*} \dfrac{d}{dx} \dfrac{x^2+4}{x-3}\\ \dfrac{d}{dx} \dfrac{\sqrt x}{3x-4}\\ \dfrac{d}{dx} \dfrac{\tan(4x)}{x} \end{align*} \]
3. Derivatives of trig functions
- The derivatives of the basic trig functions are:
\[ \begin{align*} \dfrac{d}{d \theta} \sin \theta &= \cos \theta \\ \dfrac{d}{d \theta} \cos \theta &= - \sin \theta\\ \dfrac{d}{d \theta} \tan \theta &= \sec^2 \theta\\ \dfrac{d}{d \theta} \csc \theta &= - \csc \theta \cot \theta\\ \dfrac{d}{d \theta} \sec \theta &= \sec \theta \tan \theta \\ \dfrac{d}{d \theta} \cot \theta &= - \csc^2 \theta \end{align*} \]
15. The derivatives of the inverse trigonometric functions are:
\[ \begin{align*} \frac{d}{dx} \sin^{-1}(x) &= \frac{1}{\sqrt{1-x^2}} \quad \text{for } |x| < 1, \\ \frac{d}{dx} \cos^{-1}(x) &= \frac{-1}{\sqrt{1-x^2}} \quad \text{for } |x| < 1, \\ \frac{d}{dx} \tan^{-1}(x) &= \frac{1}{1+x^2}, \\ \frac{d}{dx} \cot^{-1}(x) &= \frac{-1}{1+x^2}, \\ \frac{d}{dx} \sec^{-1}(x) &= \frac{1}{|x| \sqrt{x^2-1}} \quad \text{for } |x| > 1, \\ \frac{d}{dx} \csc^{-1}(x) &= \frac{-1}{|x| \sqrt{x^2-1}} \quad \text{for } |x| > 1 \end{align*} \]
16. Recall the following fundamental trigonometric relations and identities:
$$ \[\begin{align*} \tan \theta &= \frac{\sin \theta}{\cos \theta}, \\ \csc \theta &= \frac{1}{\sin \theta}, \\ \sec \theta &= \frac{1}{\cos \theta}, \\ \cot \theta &= \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}\\ \sin^2 \theta + \cos^2 \theta &= 1, \\ 1 + \tan^2 \theta &= \sec^2 \theta, \\ 1 + \cot^2 \theta &= \csc^2 \theta \\ \sin 2\theta &= 2 \sin \theta \cos \theta, \\ \cos 2\theta &= \cos^2 \theta - \sin^2 \theta \end{align*}\] $$
4. Derivatives of exponentials
17. The derivative of the basic exponential function is:
\[ \begin{align*}\frac{d}{dx} e^{ax} &= ae^{ax}\end{align*} \]
18. Two simple cases correspond to setting \(a\) equal to \(+1\) or \(−1\):
\[ \begin{align*}\frac{d}{dx} e^x &= e^x \\ \frac{d}{dx} e^{-x} &= -e^{-x}\end{align*} \]
19. Numerically, Euler’s number is \(e=2.718281828\) recall the following properties regarding exponentials:
\[ \begin{align*} e^{x+y} &= e^xe^y \\ e^{x-y} &= e^xe^{-y} \\ e^{-x} &= \frac{1}{e^x} \\ (e^x)^a &= e^{ax} \\ e^0 &= 1 \end{align*} \]
20. The hyperbolic functions are defined in terms of exponentials:
\[ \begin{align*} \sinh x &= \frac{e^x - e^{-x}}{2} \\ \cosh x &= \frac{e^x + e^{-x}}{2} \\ \tanh x &= \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}} \end{align*} \]
21. Note the following properties regarding hyperbolic functions:
\[ \begin{align*} \cosh^2 x - \sinh^2 x &= 1 \\ \cosh(-x) &= \cosh(x) \\ \sinh(-x) &= -\sinh(x)\\ \sinh 0 &= 0 \\ \cosh 0 &= 1 \\ \tanh 0 &= 0 \end{align*} \]
- The derivatives of the hyperbolic functions (not the ordinary trig functions) are:
\[ \begin{align*} \frac{d}{dx} \sinh x &= \cosh x \\ \frac{d}{dx} \cosh x &= \sinh x \\ \frac{d}{dx} \tanh x &= \text{sech}^2 x = \frac{1}{\cosh^2 x} \end{align*} \]
5. Derivatives of logarithms
23. The derivative of the natural logarithm function is:
\[ \begin{align*} \frac{d}{dx} \ln x &= \frac{1}{x} \end{align*} \]
24. Note that a constant multiplying the variable inside the argument has no effect on the derivative:
\[ \begin{align*} \frac{d}{dx} \ln (ax) &= \frac{1}{x} \quad \text{(the constant aa vanishes)} \end{align*} \]
25. Recall that a natural logarithm is a function used to determine the exponent in the equation \(y=e^x\). The equation \(\ln y=x\) is equivalent to \(y=e^x\)
26. Recall the following properties regarding logarithms:
\[ \begin{align*} \ln(xy) &= \ln x + \ln y \\ \ln(x/y) &= \ln x - \ln y \\ \ln(1/x) = \ln(x^{-1}) &= -\ln x \\ \ln(x^a) &= a \ln x \\ \ln(e^x) &= x \\ e^{\ln x} &= x \\ \ln(e) &= 1 \\ \ln(1) &= 0 \end{align*} \]
26. The natural logarithm is a logarithm with base Euler’s number (\(e \approx 2.718281828\)) so \(\ln x=\log_e^x\). A general logarithm of base \(b\) can be related to the natural logarithm using the change of base formula:
\[ \begin{align*}\log_b x &= \frac{\ln x}{\ln b}\end{align*} \]
27. This allows you to take a derivative of a logarithm of any base \(b\):
\[ \begin{align*} \frac{d}{dx} \log_b x = \frac{d}{dx} \left( \frac{\ln x}{\ln b} \right) = \frac{1}{\ln b} \frac{d}{dx} \ln(x) = \frac{1}{\ln b} \frac{1}{x} = \frac{1}{x \ln b} \end{align*} \]
28. A natural logarithm is also involved in the derivative of the power function:
\[ \begin{align*}\dfrac{d}{dx} b^x &= b^x \ln b, \\ \dfrac{d}{dx} b^{-x} &= -b^{-x} \ln b\end{align*} \]
29. If you set \(b=e\) in the above formulas, you get the special cases:
\[ \begin{align*}\frac{d}{dx} e^x &= e^x \\ \frac{d}{dx} e^{-x} &= -e^{-x}\end{align*} \]
which we explored in the previous chapter (note that \(\ln \e=1\)).
6. Second derivatives
30. A second derivative means to take two consecutive derivatives:
\[ \begin{align*} \frac{d^2y}{dx^2} &= \frac{d}{dx}\left(\frac{dy}{dx}\right) \end{align*} \]
7. Extreme values
31. To find the relative extrema for a function, \(f(x)\), follow these steps:
- Take the first derivative of the function, \(\frac{df}{dx}\).
- Set the first derivative equal to zero: \(\frac{df}{dx} = 0\). (Why? The slope of the tangent line is zero, corresponding to \(\frac{df}{dx} = 0\), at the relative extrema.)
- Solve for the values of \(x\) that make the first derivative zero. Call these \(x_c\) (A few books also include points where the slope is vertical or undefined.)
- Take the second derivative of the function, \(\frac{d^2f}{dx^2}\), by taking a derivative of the result from step 1.
- Evaluate the second derivative at each value of \(x_c\) found in step 3.
- For each value of \(x_c\) obtained in step 3, follow these steps to classify the type of relative extrema associated with that value of \(x_c\):
- If \(\frac{d^2f}{dx^2} > 0\) at \(x_c\), classify that point as a relative (or local) minimum.
- If \(\frac{d^2f}{dx^2} < 0\) at \(x_c\), classify that point as a relative (or local) maximum.
- If \(\frac{d^2f}{dx^2} = 0\) at \(x_c\), examine the first derivative just before and just after \(x_c\).
- Evaluate the function, \(f(x)\), at each value of \(x_c\) obtained in step 3 corresponding to a relative minimum or maximum. These are the values of the function at each relative (or local) extremum.
32. To find the absolute extrema of the function over a specified interval, \((a,b)\), evaluate the function at the endpoints of the interval, \(a\) and \(b\). Follow these steps:
- Compare the values of \(f(x_c)\) from step 7. Also compare \(f(a)\) and \(f(b)\). The largest of these values is the absolute maximum.
- Compare the values of \(f(x_c)\) from step 7. Also compare \(f(a)\) and \(f(b)\). The smallest of these values is the absolute minimum.
33. If the second derivative equals zero at \(x_c\), follow these steps:
- Determine the sign of the first derivative just before \(x_c\).
- Determine the sign of the first derivative just after \(x_c\).
- Interpret your answers to steps 1-2 as follows:
- The sign of the first derivative changes from positive to negative at a relative (or local) maximum.
- The sign of the first derivative changes from negative to positive at a relative (or local) minimum.
- If the first derivative doesn’t change sign, this is a point of inflection (where the second derivative, relating to concavity, changes sign).
8. Limits and L’hopital rule
34. The following notation asks, “What value does the function \(f(x)\) approach as the variable \(x\) approaches \(c\)?” It describes the function’s behavior as \(x\) gets closer to \(c\).
\[ \begin{align*}\lim_{x \to c} f(x)\end{align*} \]
35. A one-sided limit asks how the function behaves as \(x\) approaches \(c\) from a specified direction:
A \(+\) sign means to let \(x\) approach \(c\) from the right (start with \(x > c\) and let \(x\) decrease towards \(c\)).
A \(-\) sign means to let \(x\) approach \(c\) from the left (start with \(x < c\) and let \(x\) increase towards \(c\)).
\[ \begin{align*}\lim_{x \to c^+} f(x)\\ \lim_{x \to c^-} f(x)\end{align*} \]
36. When finding the limit of a ratio of two functions, if both approach zero or both approach infinity, l’Hôpital’s rule may apply:
\[ \begin{align*} \lim_{x \to c} \frac{f(x)}{g(x)} &= \frac{\frac{df}{dx}}{\frac{dg}{dx}} \bigg|{x=c} \quad \text{if } \lim{x \to c} f(x) = 0 \text{ and } \lim_{x \to c} g(x) = 0 \\ \lim_{x \to c} \frac{f(x)}{g(x)} &= \frac{\frac{df}{dx}}{\frac{dg}{dx}} \bigg|{x=c} \quad \text{if } \lim{x \to c} f(x) = \pm \infty \text{ and } \lim_{x \to c} g(x) = \pm \infty \end{align*} \]
9. Integrals of polynomials
38. Given a polynomial term of the form \(ax^b\) (where \(a\) is a constant coefficient and \(b\) is a constant exponent), to perform an indefinite integral over the variable \(x\), increase the exponent to \(b + 1\) and then divide by \(b + 1\), and add the constant of integration \(c\):
\[ \begin{align*} \int ax^b dx &= \frac{ax^{b+1}}{b+1} + c \quad \text{(if } b \neq -1\text{)} \end{align*} \]
39. The special case for \(b = -1\) is:
\[ \begin{align*} \int ax^{-1} dx = \int \frac{a}{x} dx &= a \ln |x| + c \end{align*} \]
40. Special exponent rules for rewriting expressions:
A variable in a denominator can be brought to the numerator by negating its exponent:
\[ \begin{align*} \frac{1}{x^n} &= x^{-n} \end{align*} \]
A square root can be rewritten using a fractional exponent:
\[ \begin{align*} \sqrt{x} &= x^{1/2} \end{align*} \]
41. If no coefficient or exponent is written, the number \(1\) is implied:
\[ \begin{align*} x &= 1x^1\\ \frac{1}{x} = \frac{1}{x^1} &= x^{-1} \end{align*} \]
42. Recall the following rules of algebra regarding exponents:
\[ \begin{align*} x^m x^n &= x^{m+n}\\ \frac{x^m}{x^n} &= x^{m-n} \\ x^0 &= 1\\ (x^m)^n &= x^{mn} \\ (ax)^m &= a^mx^m\\ \sqrt{ax} = (ax)^{1/2} &= a^{1/2}x^{1/2} \end{align*} \]
43. If a polynomial includes multiple terms, integrate each term individually:
\[ \begin{align*} \int (y_1 + y_2 + \cdots + y_n) dx &= \int y_1 dx + \int y_2 dx + \cdots + \int y_n dx \end{align*} \]
44. Example: If \(y_1 = 3x^2\), \(y_2 = -2x\), and \(y_3 = 4\), this means:
\[ \begin{align*} \int (3x^2 - 2x + 4) dx &= \int 3x^2 dx - \int 2x dx + \int 4 dx \end{align*} \]
45. Tip: After you finish an indefinite integral, take a derivative to check your answer.
10. Definite integrals
46. An indefinite integral, or antiderivative, of a function includes a constant of integration \(c\). If \(f(x)\) is the derivative of \(g(x)\), then \(g(x)\) is an antiderivative of \(f(x)\):
\[ \begin{align*}f(x) &= \frac{d}{dx}g(x) \\ \int f(x) dx &= g(x) + c\end{align*} \]
47. Example: Using \(g(x) = x^4\):
\[ \begin{align*} 4x^{3} &= \frac{d}{dx}(x^{4})\\ \int 4x^{3} dx &= x^{4} + c \end{align*} \]
An indefinite integral is the opposite of a derivative.
48. A definite integral includes limits, specifying an interval \([a, b]\):
- \(\int_{a}^{b} f(x) dx\) is a definite integral.
- \(\int f(x) dx\) is an indefinite integral.
49. The Fundamental Theorem of Calculus is used to evaluate a definite integral:
\[ \begin{align*}\int_{a}^{b} f(x) dx = g(b) - g(a)\end{align*} \]
where \(g(x)\) is any antiderivative of \(f(x)\) (i.e., \(\frac{d}{dx}g(x) = f(x)\)).
50. The procedure to evaluate a definite integral is:
- Find the antiderivative, \(g(x)\), of the integrand \(f(x)\).
- Evaluate this antiderivative at the upper limit, \(g(b)\).
- Evaluate this antiderivative at the lower limit, \(g(a)\).
- Subtract the result from step 3 from the result from step 2.
51. The constant of integration \(c\) is not needed for a definite integral because it cancels out in the subtraction:
\[ \begin{align*}g(b) + c - (g(a) + c) = g(b) - g(a)\end{align*} \]
11. Integrals of trig functions
52. The integrals of the basic trigonometric functions are:
\[ \begin{align*}\int \sin \theta d\theta &= -\cos \theta + c \\ \int \cos \theta d\theta &= \sin \theta + c \\ \int \tan \theta d\theta &= \ln |\sec \theta| + c \\ \int \sec \theta d\theta &= \ln |\sec \theta + \tan \theta| + c \\ \int \cot \theta d\theta &= \ln |\sin \theta| + c \\ \int \csc \theta d\theta &= -\ln |\csc \theta + \cot \theta| + c\end{align*} \]
53. Recall the following trigonometric relations and identities:
\[ \begin{align*} \tan \theta &= \frac{\sin \theta}{\cos \theta} \\ \csc \theta &= \frac{1}{\sin \theta} \\ \sec \theta &= \frac{1}{\cos \theta} \\ \cot \theta &= \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta} \\ \sin^2 \theta + \cos^2 \theta &= 1 \\ \tan^2 \theta + 1 &= \sec^2 \theta \\ 1 + \cot^2 \theta &= \csc^2 \theta \\ \sin 2\theta &= 2 \sin \theta \cos \theta \\ \cos 2\theta &= \cos^2 \theta - \sin^2 \theta \\ \pi \text{ radians} &= 180^\circ \end{align*} \]
12. Integrals of exponentials and logarithms
54. The integrals of the basic exponential, logarithm, and power functions are:
\[ \begin{align*}\int e^{ax} dx &= \frac{e^{ax}}{a} + c \\ \int \ln x dx &= x \ln x - x + c \\ \int b^{x} dx &= \frac{b^{x}}{\ln b} + c\end{align*} \]
55. The integrals of the hyperbolic functions are:
\[ \begin{align*}\int \sinh x dx &= \cosh x + c \\ \int \cosh x dx &= \sinh x + c \\ \int \tanh x dx &= \ln |\cosh x| + c \end{align*} \]
56. Recall the following relations and identities:
\[ \begin{align*} e &= 2.718281828...\\ e^{-x} &= \frac{1}{e^x} \\ e^{0} &= 1 \\ \ln(e) &= 1 \\ \ln(1) &= 0 \\ \ln(xy) &= \ln x + \ln y \\ \ln\left(\frac{x}{y}\right) &= \ln x - \ln y \\ \ln\left(\frac{1}{x}\right) = \ln(x^{-1}) &= -\ln x \\ \ln(x^{n}) &= n \ln x \\ \ln(e^{x}) &= x \\ e^{\ln x} &= x \\ \log_{b}x &= \frac{\ln x}{\ln b}\\ \sinh x &= \frac{e^{x}-e^{-x}}{2}\\ \cosh x &= \frac{e^{x}+e^{-x}}{2}\\ \tanh x &= \frac{\sinh x}{\cosh x} = \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \\ \cosh^{2}x - \sinh^{2}x &= 1\\ \cosh(-x) &= \cosh x \\ \sinh(-x) &= -\sinh x \\ \sinh 0 &= 0 \\ \cosh 0 &= 1 \\ \tanh 0 &= 0 \end{align*} \]
13. Integration by polynomial substitution
- A common method for performing integration is to make a change of variables (substitution). An integral over variable \(x\) can be transformed to an integral over a new variable \(u\):
\[ \begin{align*} \int_{x=x_1}^{x_2} f(x) dx = \int_{u=u_1}^{u_2} g(u) du \end{align*} \]
- To apply the method of substitution, follow these steps:
Choose a substitution \(u = u(x)\) that will simplify the integral.
Differentiate the substitution equation to relate \(du\) and \(dx\):
\[ \begin{align*}du = \frac{du}{dx} dx\end{align*} \]
Solve the above equation for \(dx\) in terms of \(du\).
Substitute \(u(x)\) and \(dx\) into the original integral.
- Replace all \(x\) terms (including in \(dx\)) with \(u\) terms.
For definite integrals: Change the limits of integration.
- The new lower limit \(u_1\) is found by evaluating \(u(x_1)\).
- The new upper limit \(u_2\) is found by evaluating \(u(x_2)\).
Evaluate the new, simpler integral in terms of \(u\).
For indefinite integrals: After integrating, substitute back to express the answer in terms of the original variable \(x\).
59. If the chosen substitution does not simplify the integral, try a different substitution or another integration method.
14. Integration by trigonometric substitution
- Trigonometric substitutions are often helpful for two types of integrals:
Integrals involving trigonometric functions:
\[ \begin{align*}\int \sin \theta \cos \theta d\theta, \quad \int \sin^3 \theta d\theta, \quad \int \frac{\sin \theta}{\cos^2 \theta} d\theta\end{align*} \]
Integrals where \(x^2\) is added or subtracted from a constant:
\[ \begin{align*} \int \sqrt{x^2 + 4} dx, \quad \int \frac{dx}{9 - x^2}, \quad \int \frac{dx}{(x^2 - 1)^{3/2}} \end{align*} \]
- Recall the following trigonometric identities and relations:
$$ \[\begin{align*} \tan \theta &= \frac{\sin \theta}{\cos \theta}\\ \csc \theta &= \frac{1}{\sin \theta} \\ \sec \theta &= \frac{1}{\cos \theta}\\ \cot \theta &= \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}\\ \sin^2 \theta + \cos^2 \theta &= 1\\ \tan^2 \theta + 1 &= \sec^2 \theta\\ 1 + \cot^2 \theta &= \csc^2 \theta \\ \sin 2\theta &= 2 \sin \theta \cos \theta \\ \cos 2\theta &= \cos^2 \theta - \sin^2 \theta \\ \sin^2 \theta &= \frac{1 - \cos 2\theta}{2} \\ \cos^2 \theta &= \frac{1 + \cos 2\theta}{2} \end{align*}\] $$
- For expressions of the form \((a^2 - x^2)^p\):
- Use the substitution: \(x = a \sin \theta\), \(dx = a \cos \theta d\theta\)
- Then: \(a^2 - x^2 = a^2 - a^2 \sin^2 \theta = a^2(1 - \sin^2 \theta) = a^2 \cos^2 \theta\)
- For expressions of the form \((x^2 + a^2)^p\):
- Use the substitution: \(x = a \tan \theta\), \(dx = a \sec^2 \theta d\theta\)
- Then: \(x^2 + a^2 = a^2 \tan^2 \theta + a^2 = a^2(\tan^2 \theta + 1) = a^2 \sec^2 \theta\)
- For expressions of the form \((x^2 - a^2)^p\):
- Use the substitution: \(x = a \sec \theta\), \(dx = a \sec \theta \tan \theta d\theta\)
- Then: \(x^2 - a^2 = a^2 \sec^2 \theta - a^2 = a^2(\sec^2 \theta - 1) = a^2 \tan^2 \theta\)
- Note: The constant \(a\) is squared in these forms.
- Example: \(\sqrt{x^2 + 4} = \sqrt{x^2 + 2^2} = (x^2 + 2^2)^{1/2}\). Here, \(a = 2\), \(p = 1/2\).
- Example: \(9 - x^2 = 3^2 - x^2\). Here, \(a = 3\), \(p = 1\).
15. Integration by parts
- Integration by parts involves using the following formula:
\[ \begin{align*} \int u dv &= uv - \int v du\\ \int_{a}^{b} u dv &= \left[ uv \right]{a}^{b} - \int{a}^{b} v du \end{align*} \]
- To apply the method of integration by parts, follow these steps:
Given an integral \(\int f(x) dx\), try to identify parts to set \(u\) and \(dv\) such that the new integral \(\int v du\) is simpler than the original.
- Choose \(u\) and \(dv\) from the integrand, where \(dv\) includes the differential \(dx\).
- Your goal is to make \(\int v du\) easier to solve than \(\int u dv\).
- When choosing \(dv\), ensure you can find its antiderivative \(v\).
- A common strategy for a product of two functions (e.g., \(\int x^2 \ln x dx\)) is to try:
- \(u = \ln x\) and \(dv = x^2 dx\), or
- \(u = x^2\) and \(dv = \ln x dx\).
- Experience and practice help in making the best choice.
Find the antiderivative of \(dv\) to get \(v\).
Find the derivative of \(u\) to get \(du\).
Substitute \(u\), \(v\), \(du\), and \(dv\) into the integration by parts formula:
\[ \begin{align*}\int u dv &= uv - \int v du\end{align*} \]
For a definite integral, remember to evaluate the term \(uv\) over the limits:
\[ \begin{align*}\int_{a}^{b} u dv &= \left[ uv \right]{a}^{b} - \int{a}^{b} v du\end{align*} \]
If the new integral \(\int v du\) is not simpler, try starting over with a different choice for \(u\) and \(dv\). Trial and error is often necessary.
16. Multiple integrals
- To perform a double or triple integral, follow these steps:
- The order of the differential elements (e.g., \(dx,dy\) vs. \(dy,dx\)) does not dictate the order of integration. The limits of integration determine the order.
- Examine the limits of integration:
- If a variable (like \(x\), \(y\), or \(z\)) appears in a limit of integration, you must perform the integral for that variable first.
- When integrating over one variable, treat all other independent variables as constants.
- For example, when integrating with respect to \(x\), treat \(y\) as a constant.
- When integrating with respect to \(y\), treat \(x\) as a constant.
- After performing an integral, evaluate its antiderivative over its limits before beginning the next integral. Work from the inside out.
- The end!
Practice Problems
Find derivative of
- \(\frac d {dx}5x^3\)
- \(\frac d {dt}t^4\)
- \(\frac{d}{dx}6x^\frac 2 3\)
- \(\frac d {du}\frac 7 u\)
- \(\frac d {dx} \sqrt{3x}\)
- \(\frac d{dx} 6x^3-12x\)
- \(\frac{d}{dx} 8x^4\)
- \(\frac{d}{dx} 5x^{-2}\)
- \(\frac{d}{dt} \frac 1 t\)
- \(\frac{d}{dx} 8x^{\frac 7 4}\)
- \(\frac{d}{dx} \frac{x^\frac 3 5}{6}\)
- \(\frac{d}{du}u\)
- \(\frac{d}{dt} \sqrt{2t}\)
- \(\frac{d}{dx}\frac{1}{\sqrt x}\)
- \(\frac{d}{dx} 5x^3+4x^2-3x+2\)
- \(\frac{d}{du}1-u\)
- \(\frac{d}{dx} 3x^\frac 3 2 + 12x^\frac 1 2\)
- \(\frac{d}{dt} \sqrt t - \frac 1 t\)
- \(\frac{d}{dx} (2x^2-5)^5\)
- \(\frac{d}{dt} \sqrt{t^3-8}\)
- \(\frac{d}{dx} (3x^4-7)(2x^3-6)\)
- \(\frac{d}{dx} \frac{x^4+3x^2-6}{x^2+1}\)
- \(\frac{d}{dx}(x^3-3x^2+4x-5)^8\)
- \(\frac{d}{dx}\frac{1}{\sqrt{5t^2-3t+6}}\)
- \(\frac{d}{dx}(4x^2-6)\sqrt x\)
- \(\frac{d}{dx} \frac{3-2x^2}{4-3x^2}\)
- \(\frac{d}{dx}\frac{1}{x^3-4x}\)
- \(\frac{d}{dx}(4+x)^9(2-x)^5\)
- \(\frac{d}{dt} \sqrt{t^2+9}\)
- \(\frac{d}{dx} \frac{x^4}{\sqrt{x^2+4}}\)
- \(\frac{d}{dx}(2x^\frac 5 2 - 8x^\frac 3 2)^6\)
- \(\frac{d}{dx} \frac{x^2+3x-4}{2x+5}\)
- \(\frac{d}{dx}\sqrt{2+\sqrt{x}}\)
- \(\frac{d}{dt}(4t^2-9)(t^4+8t^2-3)^9\)
- \(\frac{d}{d\theta}4\sin3\theta\)
- \(\frac{d}{d\theta}2\cos^3 \theta\)
- \(\frac{d}{d\theta}\sec \theta \tan \theta\)
- \(\frac{d}{d \theta} \frac{1 + \sin \theta}{\cos \theta}\)
- \(\frac{d}{dx} 4 \tan^{-1}x^3\)
- \(\frac{d}{dx} \sin^{-1} x \cos^{-1}x\), x is not equal to \(\pm1\)
- \(\frac{d}{d \theta}7 \tan 5 \theta\)
- \(\frac{d}{d\theta} 3 \sin^4 \theta\)
- \(\frac{d}{d \theta} \csc \theta \sec \theta\)
- \(\frac{d}{d \theta} \frac{\sin \theta + \cos \theta}{\sin \theta}\)
- \(\frac{d}{d \theta} \cos(\theta^2-2 \theta +4)\)
- \(\frac{d}{d \theta} 2 \cot \sqrt \theta\)
- \(\frac{d}{d\theta} \theta \sin \theta\)
- \(\frac{d}{d \theta}\sqrt{1 + \sin \theta}\)
- \(\frac{d}{dx} 3 \cot^{-1}4x\)
- \(\frac{d}{dx} x \csc^{-1}x\)
- \(\frac{d}{dx} \sin^{-1}x+ \cos^{-1}x\)
- \(\frac{d}{dx} \frac{\sec^{-1}x}{x}\)
- \(\frac{d}{dx} 3e^{x^2}\)
- \(\frac{d}{dt} t \cosh t\)
- \(\frac{d}{dx} \frac{2+e^{-2x}}{3^{3x}}\)
- \(\frac{d}{dx} 4e^{x^2} - 6e^{4x}+9\)
- \(\frac{d}{dx} 4 \cosh^3 x\)
- \(\frac{d}{dt} \sinh t \cosh t\)
- \(\frac{d}{dx} \sinh [\cosh(x)]\)
- \(\frac{d}{dx} \tanh \sqrt x\)
- \(\frac{d}{dt} t^2 e^t\)
- \(\frac{d}{dx}\sqrt{1+e^{-x}}\)
- \(\frac{d}{dx} \frac{1 - \sinh x}{\sinh x}\)
- \(\frac{d}{dx} \ln (ax)\)
- \(\frac{d}{dt} t \ln(t)\)
- \(\frac{d}{dx} \log_{10}x\)
- \(\frac{d}{dx} 2^x\)
- \(\frac{d}{dx} e^x \ln x\)
- \(\frac{d}{dt} \frac{\ln(t)}{t}\)
- \(\frac{d}{dx} \ln|\cos x|\)
- \(\frac{d}{dx} \ln(\cosh x)\)
- \(\frac{d}{dx} \sqrt{\ln(x)}\)
- \(\frac{d}{dx} \ln(\sqrt x)\)
- \(\frac{d}{dt} \log_2 t\)
- \(\frac{d}{dx} \frac{2^x}{x^2}\)
- \(\frac{d^2}{dx^2} 2x^6\)
- \(\frac{d^2}{d\theta^2} \sin \theta\)
- \(\frac{d^2}{dx^2} \frac{1}{x^2}\)
- \(\frac{d^2}{dx^2} x^7 - 3x^5 +5x^3 - 7x\)
- \(\frac{d^2}{d\theta^2} \cos 3 \theta\)
- \(\frac{d^2}{dx^2} \ln x\)
- \(\frac{d^2}{dt^2} e^{-3t^2}\)
- \(\frac{d^2}{d\theta^2} \sin(\theta^2)\)
- \(\frac{d^2}{dx^2} \tan \theta\)
- \(\frac{d^2}{dt^2} (t \ln t)\)
- \(\frac{d^2}{dx^2} \sqrt x\)
- \(\frac{d^2}{dx^2} \frac 1 x\)
- \(\frac{d^2}{d\theta^2} \sec \theta\)
- \(\frac{d^2}{dt^2} \tan^{-1} t\)
- \(\frac{d^2}{d\theta^2} (\theta \sin \theta)\)
Find absolute extrema below or over the interval
- \(f(x) = x^3-12x+4, (-3,5)\)
- \(f(x) = 2x^4-8x^3, (2,4)\)
- \(f(x) = \frac{4}{x^6} - \frac{3}{x^8}, (\frac 1 2, 10)\)
Evaluate the following limit
- \(\lim\limits_{x \to \infty}e^{-x}\)
- \(\lim\limits_{x \to \infty} \frac{6x^3-8x^2}{2x^3+4x^2}\)
- \(\lim\limits_{x \to 0} \frac{\sin x}{x}\)
- \(\lim\limits_{x \to 0} \frac{2x}{e^x-1}\)
- \(\lim\limits_{x \to 2} \frac{x^2-3x+6}{x^2+3x-2}\)
- \(\lim\limits_{x \to 0} \frac{\sqrt{144-x^2}}{\sqrt{9-x^2}}\)
- \(\lim\limits_{x \to 1} e^x \ln(x)\)
- \(\lim\limits_{x \to \pi} x \cos x\)
- \(\lim\limits_{x \to \infty} \frac{2x-8}{6x-4}\)
- \(\lim\limits_{x \to 4}\frac{x^2-16}{x-4}\)
- \(\lim\limits_{x \to \infty} \frac{2x^2-3x}{4x^2+9x}\)
- \(\lim\limits_{x \to 1} \frac{\ln x}{x-1}\)
- \(\lim\limits_{x \to 0} \frac{\tan x}{x}\)
- \(\lim\limits_{x \to \infty} \frac{2x^4 + 4x^2 +6}{x^4+2x^2+3}\)
- \(\lim\limits_{x \to \infty} \frac x {e^x}\)
- \(\lim\limits_{x \to 0} \frac{\sqrt{x+4}-2}{x}\)
- \(\lim\limits_{x \to \infty} \frac{8x^4-3x^3}{2x^6-9x^2}\)
- \(\lim\limits_{x \to 0} \frac{e^x- e^{-x}}{x}\)
- \(\lim\limits_{x \to \frac \pi 2} \frac{2x - \pi}{\cos x}\)
- \(\lim\limits_{x \to \infty} \frac{\ln(x)}{x}\)
Perform the integral
- \(\int 12x^3 ~dx\)
- \(\int \frac 6 {t^4} ~dt\)
- \(\int \frac 2 x ~dx\)
- \(\int \sqrt x ~dx\)
- \(\int (5u^4 + 9u^2)~du\)
- \(\int 35x^6 ~dx\)
- \(\int \frac{63}{t^8}~dt\)
- \(\int 15 x^{\frac 2 3}~dx\)
- \(\int 48 u^{-5}~du\)
- \(\int t ~dt\)
- \(\int \frac{dx}{\sqrt x}\)
- \(\int 4 ~du\)
- \(\int \frac{8dx}{x^\frac 1 3}\)
- \(\int (x^2 - 3x +4)~dx\)
- \(\int (\frac{1}{u^2}- \frac 4 u)~du\)
- \(\int (10x^\frac 3 2+ 6x^\frac 1 2)~dx\)
- \(\int (\sqrt t + \frac {1} {\sqrt t})~dt\)
- \(\int \limits^2_{x=1} 9x^2 ~dx\)
- \(\int \limits^{12}_{x=4} \frac{dx}{x^2}\)
- \(\int \limits^2_{x=1}8x^3~dx\)
- \(\int \limits^6_{x=3} \frac{dx}{x^3}\)
- \(\int \limits^9_{t=4} \frac{dt}{\sqrt t}\)
- \(\int \limits^2_{x=-2}(x^3 - 6x^2 + 4x -8)~dx\)
- \(\int \limits^5_{x=0} \frac{x^4}{5}~dx\)
- \(\int \limits^6_{x=-3} (t^2-2)~dt\)
- \(\int \limits^4_{x=1} \sqrt x ~dx\)
- \(\int \limits^e_{x=1} \frac{dx}{x}\)
- \(\int \limits^{\pi/2}_{\theta=0} \cos \theta ~d \theta\)
- \(\int \limits^{\pi}_{\theta=0} \sin \theta ~d \theta\)
- \(\int \limits^{\pi/3}_{\theta=0} \tan \theta ~d \theta\)
- \(\int \limits^{\pi/6}_{\theta= - \pi/6} \sec \theta ~d \theta\)
- \(\int \limits^{\pi/3}_{\theta=0}(\theta + \cos \theta) ~d \theta\)
- \(\int \limits^{1}_{x=0} \sinh x~ dx\)
- \(\int \limits^{\infty}_{x=0} e^{-x}~dx\)
- \(\int \limits^{e}_{x=1} \ln (x)~ dx\)
- \(\int \limits^{2}_{t=1} \cosh t ~dt\)
- \(\int \limits^{4}_{x=3}2^x~dx\)
- \(\int(3x+4)^5 ~dx\)
- \(\int \limits^{\pi / 8}_{\theta=0} \cos(4 \theta)~ d \theta\)
- \(\int \limits^{2}_{x=1} x^2 \sqrt{x^3 + 8}~dx\)
- \(\int \frac{dx}{2x+3}\)
- \(\int \limits^{\pi/9}_{\theta=0} 6 \sin(3 \theta) ~d \theta\)
- \(\int \frac{dx}{\sqrt{x-1}}\)
- \(\int \limits^{\infty}_{t=0} t e^{-t^2}~dt\)
- \(\int \frac{8x^3}{x^4-2}~dx\)
- \(\int \limits^{\sqrt \pi/2}_{\theta=0} \theta \cos(\theta^2)~d \theta\)
- \(\int \frac{12z^2}{(x^3+16)^2}~dx\)
- \(\int \limits^{\pi^2/4}_{\theta=0} \frac{\sin(\sqrt \theta)}{\sqrt \theta}~d \theta\)
- \(\int \sqrt{1- \sqrt x}~dx\)
- \(\int \limits^{4}_{x=0} \sqrt{3x+4}~dx\)
- \(\int \sin^3 \theta \cos \theta ~d \theta\)
- \(\int \sin^2 \theta ~d \theta\)
- \(\int \limits^{\pi/2}_{\theta=0} \cos^3 \theta ~d \theta\)
- \(\int \limits^{3}_{x=0} \frac{dx}{\sqrt{x^2+9}}\)
- \(\int \frac{dx}{\sqrt{16-x^2}}\)
- \(\int \limits^{\pi/3}_{\theta=0} \cos^4 \theta \sin \theta ~d\theta\)
- \(\int \frac{dx}{x^2+25}\)
- \(\int \limits^{\pi/4}_{\theta=0} \tan^3 \theta ~d \theta\)
- \(\int \sin^4 \theta ~d \theta\)
- \(\int \limits^{1}_{x=0} \frac{dx}{(x^2+1)^2}\)
- \(\int \frac{dx}{(9-x^2)^\frac 3 2}\)
- \(\int \limits^{\pi/2}_{\theta=\pi/6} \frac{\cos \theta}{\sin^3 \theta} d\theta\)
- \(\int x \sin x ~dx\)
- \(\int \limits^{e}_{x=1} x^3 \ln(x)~dx\)
- \(\int x e^x ~dx\)
- \(\int \limits^{\pi/2}_{x=0} x \cos x ~dx\)
- \(\int \frac{\ln(x)}{x^2}dx\)
- \(\int \limits^{\pi/6}_{x=0} \sin x \tan x ~dx\)
- \(\int x^2 \cos x ~dx\)
- \(\int e^x \sin x ~dx\)
- \(\int \limits^{2}_{x=0} \int \limits^{x^2}_{y=0} xy ~dx dy\)
- \(\int \limits^{1/y}_{x=0} \int \limits^{3}_{y=1} xy^2 ~dxdy\)
- \(\int \limits^{1/3}_{x=0} \int \limits^{3}_{y=0} x^2 y^2 ~dxdy\)
- \(\int \limits^{1}_{x=0} \int \limits^{x^2}_{y=0} \int \limits^{y}_{z=0} xyz ~dxdydz\)
- \(\int \limits^{3}_{x=0} \int \limits^{\sqrt x}_{y=0} xy ~dxdy\)
- \(\int \limits^{y^2}_{x=0} \int \limits^{2}_{y=1} \frac{y^2}{\sqrt x} dxdy\)
- \(\int \limits^{4}_{x=1} \int \limits^{9}_{y=4} \frac{dxdy}{\sqrt{xy}}\)
- \(\int \limits^{5}_{x=0}\int \limits^{2x}_{y=x} x^2y ~dxdy\)
- \(\int \limits^{y}_{x=0}\int \limits^{2}_{y=0}\int \limits^{x}_{z=0} xy^2z^3 ~dxdydz\)
- \(\int \limits^{\sqrt y}_{x=0}\int \limits^{\sqrt z}_{y=0}\int \limits^{4}_{z=0} x^3y~ dxdydz\)