Calculus with Dr. Bob III: Log, Exponential and Inverse Trig Functions

Video Lectures

Displaying all 48 video lectures.
Lecture 1
Definition of ln(x) Using Integration
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Definition of ln(x) Using Integration
Calculus: We define the natural logarithm as an antiderivative of 1/x using the Second Fundamental Theorem of Calculus. We investigate the graph and explain how to estimate values of ln(x). ADDED: Some methods for approximating e using calculus.
Lecture 2
Properties of ln(x) Using Integration
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Properties of ln(x) Using Integration
Calculus: We derive two key properties for ln(x) as a definite integral: the exponent rule and the product rule. We provide several methods for each property, including u-substitution and uniqueness of antiderivatives up to a constant.
Lecture 3
End Behavior of ln(x)
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End Behavior of ln(x)
Calculus: Using the exponent rule for natural logarithm, we verify the end behavior of the graph of ln(x). That is, we show that the limit as x goes to infinity of ln(x) is positive infinity and that the limit as x goes to 0 from the right is negative infinity. We interpret the result as an area.
Lecture 4
Derivative of ln(x) as a Slope
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Derivative of ln(x) as a Slope
Calculus: With the properties developed so far, we verify the derivative of ln(x) by using the definition as the slope of a tangent line. A special limit is required, which we verify at a few points.
Lecture 5
The Chain Rule for ln(x) and ln|x|
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The Chain Rule for ln(x) and ln|x|
Calculus: We consider derivatives for functions of the form ln(g(x)) and ln|g(x)|. As preparation, we note some details concerning the domain of ln(x). Example include y(x) = ln|x^2-4x| and y(x) = ln|ln(x)|.
Lecture 6
Derivative of f(x) = ln(ln(x^2 + 1))
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Derivative of f(x) = ln(ln(x^2 + 1))
Worked problem in calculus. The derivative of f(x) = ln(ln(x^2 + 1)) is calculated using the chain rule twice.
Lecture 7
Graph of f(x) = 3ln(x-2) + 4
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Graph of f(x) = 3ln(x-2) + 4
Worked problem in calculus. The graph of 3ln(x-2) + 4 is derived from the graph of ln(x).
Lecture 8
Graph of f(x) = ln(x^2 - x)
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Graph of f(x) = ln(x^2 - x)
Worked problem in calculus. Sketch the graph of ln(x^2-x). Regions of increase/decrease and concavity are determined.
Lecture 9
Graph of f(x) = ln|x^2-x|
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Graph of f(x) = ln|x^2-x|
Calculus: Using the chain rule for ln|x|, we sketch the graph of f(x) = ln|x^2-x|. Steps include finding the domain, regions of increasing-decreasing and concavity, and end behavior.
Lecture 10
Implicit Differentiation with ln(x)
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Implicit Differentiation with ln(x)
Worked problem in calculus. Find the tangent line to the curve xy + ln(xy^2) = 1 at the point (1, 1). Implicit differentiation is used to find the slope of the tangent line.
Lecture 11
Logarithmic Differentiation
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Logarithmic Differentiation
Worked problem in calculus. Find the derivative f'(x) of the function f(x) = (x+1)^(1/3)/[x^2(x+2)^(1/5)] using logarithmic differentiation. Logarithmic differentiation is a useful alternative to the quotient rule for derivatives.
Lecture 12
Antiderivative of ln(x^4)/x
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Antiderivative of ln(x^4)/x
Calculus: The antiderivative of ln(x^4)/x is computed in two ways, both using integration by substitution. The main concept is that the derivative of ln|x| is 1/x. Care must be taken when simplifying with the exponent rule for logarithm.
Lecture 13
Antiderivative involving 1/x 1
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Antiderivative involving 1/x 1
Worked problem in calculus. The antiderivative of f(x) = (x^3-2x^2+x+1)/(x+2) is computed by two methods. The first uses synthetic division. The second uses the substitution u = x+2. In both cases, a key step is that int 1/x dx = ln|x| + C
Lecture 14
Antiderivative involving 1/x 2
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Antiderivative involving 1/x 2
Worked problem in calculus. We calculate the indefinite integrals of the functions
(a) f(x) = 1/[x^(1/3) (1+x^(2/3))] and (b) f(x) = x(x-3)/(x + 2)^3. A key step uses that int 1/x = ln|x| + c.
Lecture 15
Trig Antiderivatives involving 1/x
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Trig Antiderivatives involving 1/x
Worked problem in calculus. Find the indefinite integral of (a) f(x) = 2x sec(1+x^2) and (b) f(x) = tan(ln(x))/x. A key step is that int 1/x dx = ln|x| + c.
Lecture 16
Area under f(x) = ln(x)/x
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Area under f(x) = ln(x)/x
Worked problem in calculus. The area under the curve f(x) = ln(x)/x between x=1 and x=e above the x-axis is calculated. A key step is that int 1/x dx = ln|x| + c.
Lecture 17
Inverse Functions
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Inverse Functions
Worked problem in calculus. Determine if the following functions have inverses. If so, find the inverse function, check, and sketch. (a) f(x) = (x-3)^3 + 4, (b) f(x) = |x-2| + |x-1|, and (c) f(x) = 2sqrt(x - 3) + 1.
Lecture 18
Inverse Functions: ln(x) and exp(x)
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Inverse Functions: ln(x) and exp(x)
Calculus: Because ln(x) is a one-one function, we consider its inverse function exp(x). We interpret the usual properties for inverse functions with these functions.
Lecture 19
Graph of f(x) = 2 - e^{2x-1}
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Graph of f(x) = 2 - e^{2x-1}
Worked problem in calculus. The graph of f(x) = 2- e^(2x-1) is derived from the graph of g(x) = e^x.
Lecture 20
Inverse Function for f(x) = 2 e^{x-2}
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Inverse Function for f(x) = 2 e^{x-2}
Worked problem in calculus. For the function f(x) = 2 e^(x-2), the inverse is found and checked. Then the derivative rule for inverse functions is verified.
Lecture 21
No Inverse Function for f(x)=ln(x)-exp(x)
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No Inverse Function for f(x)=ln(x)-exp(x)
Calculus: Let f(x) = ln(x) - exp(x). We show that f(x) has no inverse function by considering 1) end behavior and 2) regions of increasing an decreasing. By restricting the domain, we find an inverse function and approximate f^{-1}(1) and f^{-1}(-3).
Lecture 22
Derivative of an Inverse Function
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Derivative of an Inverse Function
Worked problem in calculus. Find the inverse function, check, and verify the derivative rule for the functions (a) f(x) = 3x-2, and (b) g(x) = sqrt(x-3).
Lecture 23
Properties of Derivative of exp(x)
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Properties of Derivative of exp(x)
Calculus: We show that the derivative of exp(x) is exp(x). We verify derivative properties of the graph of exp(x), show consistency with the slope definition of derivative, and note how the chain rule applies.
Lecture 24
Derivatives with e^x
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Derivatives with e^x
Worked problem in calculus. The derivative is found for the following functions: (a) f(x) = x^2 e^{-x+ln(x)}, (b) g(x) = ln(1+e^{x^2}), and (c) h(x) = int_0^ln(x) sqrt(1+e^t) dt. Part (c) is an application of the Second Fundamental Theorem of Calculus.
Lecture 25
Tangent Line to x^2 e^x at x=1
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Tangent Line to x^2 e^x at x=1
Worked problem in calculus. The formula for the tangent line to f(x) = x^2 e^{-x} at x=1 is derived. The line is used to approximate f(1.1).
Lecture 26
Graph of f(x) = x^2 e^{-x}
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Graph of f(x) = x^2 e^{-x}
Worked problem in calculus. The graph of f(x) = x^2 e^{-x} is derived from the first and second derivatives.
Lecture 27
Graph of Normal Distribution
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Graph of Normal Distribution
Calculus: As an application of the chain rule with exp(x), we sketch the function f(x) = exp(-(x-m)^2/(2s^2)), a multiple of a normal distribution. Using the first and second derivatives, we note critical and inflection points, and regions of increasing, decreasing, and concavity.
Lecture 28
Implicit Differentiation with e^x
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Implicit Differentiation with e^x
Calculus: Consider the curve defined by exp(xy) + x^2 + y^2 = 5. Use implicit differentiation to find the tangent line to the curve at (2, 0). Then use the tangent line to approximate the point near (2, 0) with x = 2.01.
Lecture 29
Indefinite Integrals with e^x
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Indefinite Integrals with e^x
Worked problem in calculus. Evaluate the indefinite integrals of (a) e^{-3x+4}, (b) (e^x - 2)(e^x -3)/e^x and (c) e^x/ (1 + 2e^x).
Lecture 30
Definite Integral of exp(-3x+2) (HD Version)
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Definite Integral of exp(-3x+2) (HD Version)
Calculus: Compute the definite integral of exp(-3x+2) over the interval [0, ln(2)]. We apply integration by substitution to solve.
Lecture 31
Definite Integral of (e^x - 1)(e^x - 2)/e^2x (HD Version)
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Definite Integral of (e^x - 1)(e^x - 2)/e^2x (HD Version)
Calculus: Compute the definite integral of (e^x - 1)(e^x - 2)/e^2x over the interval [ln(2), ln(3)]. We use integration by substitution to solve.
Lecture 32
Graphing with 2^x
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Graphing with 2^x
Worked problem in calculus. Graph the function f(x) = 3.2^(x-1) - 4.
Lecture 33
Equations with Logarithms
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Equations with Logarithms
Worked problem in calculus. The equations log_6(x-5) + log_6(x) = 1 and log_b(9) - 3log_b(3) = 1/2 are solved.
Lecture 34
Derivatives of log_b(x) and b^x
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Derivatives of log_b(x) and b^x
Worked problem in calculus. Find the derivative of f(x)= x.2^(3x+1) and g(x) = log_2(x^2-2x-3/x).
Lecture 35
Tangent Lines to log_b(x) and b^x
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Tangent Lines to log_b(x) and b^x
Worked problem in calculus. Compute the tangent lines to the following functions at x=1 and approximate at x=1.1. (a) f(x) = 3^(x^2-x), (b) f(x) = log_2(x+1).
Lecture 36
Logarithmic Differentiation
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Logarithmic Differentiation
Worked problem in calculus. Logarithmic differentiation is used to compute the derivative of f(x) = (x^2-x)^(2x+1).
Lecture 37
Indefinite Integrals with b^x and log_b(x)
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Indefinite Integrals with b^x and log_b(x)
Worked problem in calculus. Find the indefinite integrals (a) (2x-1) int 2^(x^2-x) dx and (b) int log_2(x)/x dx.
Lecture 38
Definite Integrals with b^x and log_b(x)
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Definite Integrals with b^x and log_b(x)
Worked problem in calculus. The following definite integrals are computed: (a) int_1^2 2^(3x-1) dx, and (b) int_1^2 [log_2(x)]^3/x dx.
Lecture 39
Compound Interest
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Compound Interest
Worked problem in calculus. Given an initial investment of $1000 and a nominal annual interest rate of 5%, compute the amount accrued after 10 years by compounding annually, monthly, daily, and continuously. Find the time to double the initial investment if compounding is annual or continuous.
Lecture 40
Evaluating Inverse Trig Functions 1
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Evaluating Inverse Trig Functions 1
Worked problem in calculus. We evaluate the inverse trig functions: (a) sin^{-1}(sqrt(3)/2), (b) cos^{-1}(3), (c) tan^{-1}(sqrt(3)/3), and (d) sec^{-1}(sqrt(2)).
Lecture 41
Evaluating Inverse Trig Expressions 2
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Evaluating Inverse Trig Expressions 2
Worked problem in calculus. Evaluate (a) cos(sin^{-1}(3/5)), (b) tan(sin^{-1}(-3/5)), and (c) csc(sec^{-1}(sqrt(2))
Lecture 42
Evaluating Inverse Trig Expressions 3
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Evaluating Inverse Trig Expressions 3
Worked problem in calculus. Find the algebraic expressions for (a) cos(sin^{-1}(x/(x+1)), (b) sin(tan^{-1}(x)), and (c) csc(sec^{-1} (x)).
Lecture 43
Derivatives of Inverse Trig Functions
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Derivatives of Inverse Trig Functions
Worked problem in calculus. Find the derivatives of (a) g(x) = xsin^{-1}(x^2+x), (b) h(x) = tan^{-1}(ln(x)), and (c) s(x) = sec^{-1}(e^2x).
Lecture 44
Graph of f(x) = tan^{-1}(1-x^2)
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Graph of f(x) = tan^{-1}(1-x^2)
Worked problem in calculus. Assuming the graph has a horizontal asymptote at y = pi/2, sketch the graph of f(x) = tan^{-1}(1-x^2) using the first and second derivatives. We make use of the quadratic equation for ax^4 + bx^2 + c with y=x^2.
Lecture 45
Tangent Line to f(x) = xsin^{-1}(x/2)
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Tangent Line to f(x) = xsin^{-1}(x/2)
Worked problem in calculus. We compute the equation of the tangent line to f(x) = xsin^{-1}(x/2) at x=1. Using the tangent line, we approximate f(1.1).
Lecture 46
Integrals with Inverse Trig Functions 1
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Integrals with Inverse Trig Functions 1
Worked problem in calculus. The indefinite integrals int 1/[4 + (x+2)^2] dx and
int dx/sqrt(-4x-x^2) are calculated. The first integral uses the rule int du/(1+u^2) = tan^(-1) (u) + C. The second integral uses completing the square and the rule int du/sqrt(1-u^2).
Lecture 47
Integrals with Inverse Trig Functions 2
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Integrals with Inverse Trig Functions 2
Worked problem in calculus. The definite integrals int_(-3)^0 dx/(x^2+6x+18), int_(-3)^0 (2x+6)dx/(x^2+6x+18), and int_(-3)^0 (2x+8)dx/(x^2+6x+18) are computed. The different approaches to each integral are noted.
Lecture 48
Integrals with Inverse Trig Functions 3
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Integrals with Inverse Trig Functions 3
Worked problem in calculus. We calculate the definite integrals int_1^2 dx/sqrt(4-x^2) and int_0^(sqrt(3)/2) sin^-1 (x) dx/sqrt(1-x^2). The first integral uses the rule int dx/sqrt(1-x^2) = sin^-1(x) + C. The second rule is the special u-substitution when we have f(x)f'(x) as integrand.