17Calculus - Logarithmic Differentiation
This topic is usually found in the section discussing implicit differentiation and sometimes instructors do not make a distinction between the two. But logarithmic differentiation is a very specific technique and often uses implicit differentiation along the way.
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There are two main types of equations that you will use logarithmic differentiation on
1. equations where you have a variable in an exponent
2. equations that are quite complicated and can be simplified using logarithms.
In both cases, we introduce logarithms into the equation that may not have been there before, apply some simple rules and then take the derivative. Let's look at each case.
Variables In The Exponent
Remember that you can use the power rule on \(x^2\) but you can't use the power rule on \(2^x\) or \(y^x\). |
So, what we do is introduce a natural log into the equation, without changing the problem of course. The goal is to bring the exponent down so that we can take the derivative of it.
These are the rules we use |
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1. \( \ln(x^y) = y~ \ln(x) \) |
2. \( e^{\ln(z)} = z \) |
Notice the first rule brings the exponent down in front of the natural log term, in which case, we can use the product rule to take the derivative. The second rule is usually used to reverse the process after taking the derivative.
Let's look at an example. One of the practice problems shows how to calculate the derivative of \( y = x^x \). To do this one, we need to bring the x in the exponent down (since we can't use the power rule). To accomplish this, we take the natural log of both sides, like this:
\(\displaystyle{
\begin{array}{rcl}
y & =& x^x \\
\ln(y) & = & \ln(x^x) \\
\ln(y) & = & x \ln(x)
\end{array}
}\)
Now we can take the derivative of this last equation using chain rule and the product rule. [For a complete solution, see practice problem 890.]
Simplifying
In some cases, we could use the product and/or quotient rules to take a derivative but, using logarithmic differentiation, the derivative would be much easier to find. One of the practice problems is to take the derivative of \(\displaystyle{ y = \frac{(\sin(x))^2(x^3+1)^4}{(x+3)^8} }\). We could use the product and quotient rules here but, if we take the logarithm of both sides, simplify, take the derivative, then convert back, it is much easier. [see practice problem 892 for how to find the derivative \( dy/dx \)]
Practice
Unless otherwise instructed, use logarithmic differentiation to calculate the derivative of these functions.
Basic |
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\( y=5^x \)
Problem Statement |
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Use logarithmic differentiation to calculate the derivative \(dy/dx\) of \( y = 5^x \)
Solution |
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1543 video
video by MathTV |
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\( y = x^x \)
Problem Statement |
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Use logarithmic differentiation to calculate the derivative \(dy/dx\) of the function \( y = x^x \)
Final Answer |
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\( d[x^x]/dx = x^x(1+\ln(x)) \)
Problem Statement |
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Use logarithmic differentiation to calculate the derivative \(dy/dx\) of the function \( y = x^x \)
Solution |
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\( \ln(y) = \ln(x^x) \) |
\( \ln(y) = x \ln(x) \) |
\(\displaystyle{ \frac{d}{dx}[\ln(y)] = \frac{d}{dx}[x \ln(x)] }\) |
\( (1/y)(dy/dx) = x(1/x) + \ln(x)(1) \) |
\( dy/dx = y(1+\ln(x)) \) |
\( dy/dx = x^x(1+\ln(x)) \) |
Notice we could not use the power rule since the exponent is not a rational number.
Here is a video solution.
890 video
Final Answer |
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\( d[x^x]/dx = x^x(1+\ln(x)) \) |
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For \( f(x)=x^{2x} \), determine \(f'(x)\) and \(f'(1)\) using logarithmic differentiation.
Problem Statement |
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For \( f(x)=x^{2x} \), determine \(f'(x)\) and \(f'(1)\) using logarithmic differentiation. Give your answers in exact, completely factored form.
Solution |
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1546 video
video by MIP4U |
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\( y = x^{e^x} \)
Problem Statement |
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Use logarithmic differentiation to calculate the derivative \(dy/dx\) of the function \( y = x^{e^x} \)
Final Answer |
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\( dy/dx=x^{e^x-1}e^x(1+x\ln(x)) \)
Problem Statement |
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Use logarithmic differentiation to calculate the derivative \(dy/dx\) of the function \( y = x^{e^x} \)
Solution |
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Notice we cannot use the power rule since the exponent, \(e^x\), is not a rational number. So to get the factor out of the exponent, we use logarithms.
\( \ln(y) = \ln(x^{e^x}) \) |
\( \ln(y) = e^x \ln(x) \) |
\(\displaystyle{ \frac{1}{y} \cdot \frac{dy}{dx} = e^x (1/x) + \ln(x) e^x }\) |
\(\displaystyle{ \frac{dy}{dx} = ye^x(1/x+\ln(x)) }\) |
\(\displaystyle{ \frac{dy}{dx} = \frac{ye^x}{x}(1+x \ln(x)) }\) |
\(\displaystyle{ \frac{dy}{dx} = \frac{x^{e^x}e^x}{x}(1+x \ln(x)) }\) |
\(\displaystyle{ \frac{dy}{dx} = x^{e^x-1}e^x(1+x \ln(x)) }\) |
Final Answer |
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\( dy/dx=x^{e^x-1}e^x(1+x\ln(x)) \) |
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\( y=(\ln x)^x \)
Problem Statement |
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Use logarithmic differentiation to calculate the derivative \(dy/dx\) of the function \( y=(\ln x)^x \)
Solution |
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894 video
video by PatrickJMT |
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\( y = x^{\sin(x)} \)
Problem Statement |
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Use logarithmic differentiation to calculate the derivative \(dy/dx\) of the function \( y = x^{\sin(x)} \)
Solution |
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1544 video
video by MathTV |
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\(y=(x^2)^{\sin(x)}\)
Problem Statement |
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Use logarithmic differentiation to calculate the derivative \(dy/dx\) of the function \(y=(x^2)^{\sin(x)}\)
Solution |
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895 video
video by PatrickJMT |
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\(\displaystyle{y=\frac{(x+2)^2}{\sqrt{x^2+1}}}\)
Problem Statement |
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Use logarithmic differentiation to calculate the derivative \(dy/dx\) of the function \(\displaystyle{y=\frac{(x+2)^2}{\sqrt{x^2+1}}}\)
Solution |
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1545 video
video by MIP4U |
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\(\displaystyle{y=\frac{x^4\sqrt{x-3}}{(x+1)^3}}\)
Problem Statement |
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Use logarithmic differentiation to calculate the derivative \(dy/dx\) of the function \(\displaystyle{y=\frac{x^4\sqrt{x-3}}{(x+1)^3}}\)
Solution |
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1548 video
video by MIP4U |
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\(\displaystyle{ y = \sqrt{\frac{1-x}{1+x}} }\)
Problem Statement |
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Use logarithmic differentiation to calculate the derivative \(dy/dx\) of the function \(\displaystyle{ y = \sqrt{\frac{1-x}{1+x}} }\)
Solution |
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1550 video
video by Houston Math Prep |
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\( y=x^{x^2} \)
Problem Statement |
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Use logarithmic differentiation to calculate the derivative \(dy/dx\) of the function \( y=x^{x^2} \)
Solution |
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1552 video
video by Houston Math Prep |
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\( y = (\sin(x))^x \)
Problem Statement |
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Use logarithmic differentiation to calculate the derivative \(dy/dx\) of the function \( y = (\sin(x))^x \)
Solution |
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1553 video
video by TheInfiniteLooper |
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\( y = (\cos x)^{\tan x} \)
Problem Statement |
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Use logarithmic differentiation to calculate the derivative \(dy/dx\) of the function \( y = (\cos x)^{\tan x} \)
Solution |
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1554 video
video by TheInfiniteLooper |
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\(\displaystyle{ y=\frac{2(x^2+1)}{\sqrt{\cos(2x)}} }\)
Problem Statement |
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Use logarithmic differentiation to calculate the derivative \(dy/dx\) of the function \(\displaystyle{ y=\frac{2(x^2+1)}{\sqrt{\cos(2x)}} }\)
Final Answer |
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\(\displaystyle{ \frac{dy}{dx} = \left[ \frac{2x}{x^2+1} + \tan(2x) \right] \frac{2(x^2+1)}{\sqrt{\cos(2x)}} }\)
Problem Statement |
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Use logarithmic differentiation to calculate the derivative \(dy/dx\) of the function \(\displaystyle{ y=\frac{2(x^2+1)}{\sqrt{\cos(2x)}} }\)
Solution |
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There is a mistake in this video. When he first takes the derivative, he drops a factor of two from the first term on the right side of the equal sign. Actually, I would do one more simplification than he does before taking the derivative by expanding \(\ln [2(x^2+1)] = \ln 2 + \ln(x^2+1)\). Then when you take the derivative of \(\ln 2\) you get zero.
1555 video
video by SchoolOfChuck |
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Final Answer |
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\(\displaystyle{ \frac{dy}{dx} = \left[ \frac{2x}{x^2+1} + \tan(2x) \right] \frac{2(x^2+1)}{\sqrt{\cos(2x)}} }\) |
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\(\displaystyle{ y=(\sin \theta)^{\sqrt{\theta}} }\)
Problem Statement |
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Use logarithmic differentiation to calculate the derivative \( dy/d\theta \) of the function \(\displaystyle{ y=(\sin \theta)^{\sqrt{\theta}} }\)
Solution |
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1556 video
video by SchoolOfChuck |
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\(\displaystyle{ f(x)=(2x-3)^2(5x^2+2)^3 }\)
Problem Statement |
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Use logarithmic differentiation to calculate the derivative \(dy/dx\) of the function \(\displaystyle{ f(x)=(2x-3)^2(5x^2+2)^3 }\)
Solution |
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1549 video
video by MIP4U |
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Intermediate |
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Find all values of \(x\) for which the curve \( y = (x^2/4)^x \) has horizontal tangents.
Problem Statement |
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Find all values of \(x\) for which the curve \( y = (x^2/4)^x \) has horizontal tangents.
Final Answer |
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\( x = \pm 2/e \)
Problem Statement |
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Find all values of \(x\) for which the curve \( y = (x^2/4)^x \) has horizontal tangents.
Solution |
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\(\displaystyle{ y = (x^2/4)^x }\) |
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\(\displaystyle{ \ln(y) = \ln[ (x^2/4)^x ] }\) |
\(\displaystyle{ \ln(y) = x\ln(x^2/4) }\) |
\(\displaystyle{ \ln(y) = x[ \ln x^2 - \ln 4] }\) |
\(\displaystyle{ \ln(y) = x[ 2\ln x - \ln 4] }\) |
\(\displaystyle{ \ln(y) = [ 2x\ln x - x\ln 4] }\) |
\(\displaystyle{ \frac{d}{dx}[\ln(y)] = \frac{d}{dx}[ 2x\ln x - x\ln 4] }\) |
\(\displaystyle{ \frac{1}{y}\frac{dy}{dx} = (2x)(1/x) + 2\ln(x) - \ln 4 }\) |
\(\displaystyle{ \frac{dy}{dx} = y[ 2 + \ln(x^2/4) ] }\) |
\(\displaystyle{ \frac{dy}{dx} = (x^2/4)^x [ 2 + \ln(x^2/4) ] }\) |
Plotting the \( y = (x^2/4)^x \), which is the first term on the right of the equal sign, we can see that this function is never equal to zero. Using the zero-product rule, we can solve for \(x\) in the second term. |
\(\displaystyle{ 2 + \ln(x^2/4) = 0 }\) |
\(\displaystyle{ \ln(x^2/4) = -2 }\) |
\(\displaystyle{ x^2/4 = e^{-2} }\) |
\(\displaystyle{ x^2 = 4e^{-2} }\) |
\(\displaystyle{ x = \pm 2/e }\) |
Although not asked to in the problem statement, we decided to plot the original function and check our answers.
Final Answer |
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\( x = \pm 2/e \) |
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\(\displaystyle{y=\frac{(\sin(x))^2(x^3+1)^4}{(x+3)^8}}\)
Problem Statement |
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Use logarithmic differentiation to calculate the derivative \(dy/dx\) of the function \(\displaystyle{y=\frac{(\sin(x))^2(x^3+1)^4}{(x+3)^8}}\)
Solution |
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892 video
video by PatrickJMT |
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\(\displaystyle{ y=\sqrt{x}e^{x^2}(x^2+1)^{10} }\)
Problem Statement |
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Use logarithmic differentiation to calculate the derivative \(dy/dx\) of the function \(\displaystyle{ y=\sqrt{x}e^{x^2}(x^2+1)^{10} }\)
Solution |
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893 video
video by PatrickJMT |
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For \(\displaystyle{f(x)=\frac{x^2(x+2)^4}{(2x^2-1)^3}}\), find \(f'(x)\) and \(f'(4)\) using logarithmic differentiation.
Problem Statement |
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For \(\displaystyle{f(x)=\frac{x^2(x+2)^4}{(2x^2-1)^3}}\), find \(f'(x)\) and \(f'(4)\) using logarithmic differentiation. Give your answers in exact, completely factored form.
Solution |
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1547 video
video by MIP4U |
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\(\displaystyle{y=\sqrt[3]{\frac{x(x^2+1)^4}{x^3-2}}}\)
Problem Statement |
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Use logarithmic differentiation to calculate the derivative \(dy/dx\) of the function \(\displaystyle{y=\sqrt[3]{\frac{x(x^2+1)^4}{x^3-2}}}\)
Solution |
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1551 video
video by Houston Math Prep |
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Advanced |
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\(\displaystyle{y=x^{(x^x)}}\)
Problem Statement |
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Use logarithmic differentiation to calculate the derivative \(dy/dx\) of the function \(\displaystyle{y=x^{(x^x)}}\)
Solution |
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896 video
video by Khan Academy |
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You CAN Ace Calculus
Topics You Need To Understand For This Page
Related Topics and Links
external links you may find helpful |
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WikiBooks - Derivatives of Exponential and Logarithm Functions |
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
|
Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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Practice Instructions
Unless otherwise instructed, use logarithmic differentiation to calculate the derivative of these functions.
