17Calculus Derivatives - Logarithm Functions
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The logarithm rule is not as simple as the exponential derivative but it is still very straightforward.
Basic Logarithm Rule |
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\(\displaystyle{ \frac{d}{dt}[\ln(t)] = \frac{1}{t} }\) |
Logarithm With Chain Rule |
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\(\displaystyle{ \frac{d}{dt}[\ln(u)] = \frac{1}{u}\frac{du}{dt} }\) |
It is probably not clear just from the equation that the derivative of \(\ln(x)\) is \(1/x\). Here is a great video explaining, first intuitively, then from the limit, where this derivative comes from. Although he says you can stop the video after the intuitive explanation, watching the entire video will help you a lot (it's not very long).
MathTV - The Derivative of the Natural Log Function [12min-15secs]
video by MathTV |
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Many times, it helps to simplify a logarithmic expression before taking the derivative. Here are a few rules that should help you.
\( \ln(xy) = \ln(x) + \ln(y) \) |
\( \ln(x/y) = \ln(x) - \ln(y) \) |
\( \ln(x^y) = y \ln(x) \) |
Here is a short video clip that goes through these equations again.
PatrickJMT - Derivatives of Logarithmic Functions and Examples [1min-29secs]
video by PatrickJMT |
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Before jumping into some practice problems, take a couple of minutes to watch this next video. It will help you see some common mistakes that you can avoid when taking the derivative of logarithm functions.
MathTV - Common Mistakes for Natural Logs and the Chain Rule [3min-43secs]
video by MathTV |
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Practice
Unless otherwise instructed, calculate the derivative of these functions.
Here are a few practice problems that do not require the chain rule.
\(\displaystyle{f(x)=\frac{1}{\ln(x)}}\)
Problem Statement |
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Calculate the derivative of \(\displaystyle{f(x)=\frac{1}{\ln(x)}}\)
Solution |
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1071 video
video by Krista King Math |
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\(f(x)=\sqrt{x}\ln(x)\)
Problem Statement |
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Calculate the derivative of \(f(x)=\sqrt{x}\ln(x)\)
Solution |
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1085 video
video by PatrickJMT |
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\(\displaystyle{y=\frac{\ln x}{1+\ln(2x)}}\)
Problem Statement |
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Calculate the derivative of \(\displaystyle{y=\frac{\ln x}{1+\ln(2x)}}\)
Solution |
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Although the chain is not required here, the instructor in this video does use the chain rule to calculate the derivative. You can use a logarithm rule to get \(\ln(2x) = \ln(2)+\ln(x)\) before taking the derivative, in order to avoid the chain rule.
1091 video
video by PatrickJMT |
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\(y=\ln(x)+2^x+\sin(x)\)
Problem Statement |
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Calculate the derivative of \(y=\ln(x)+2^x+\sin(x)\)
Solution |
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1080 video
video by PatrickJMT |
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These practice problems require the chain rule.
Basic |
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\(f(x)=\ln(x^2+10)\)
Problem Statement |
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Calculate the derivative of \(f(x)=\ln(x^2+10)\)
Solution |
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1084 video
video by PatrickJMT |
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\(y=\ln(x^2+x)\)
Problem Statement |
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Calculate the derivative of \(y=\ln(x^2+x)\)
Solution |
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1088 video
video by PatrickJMT |
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\(\displaystyle{ \ln(3x^2+9x-5) }\)
Problem Statement |
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Calculate the derivative of this function and give your final answer in completely factored form. \(\displaystyle{ \ln(3x^2+9x-5) }\)
Final Answer |
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\(\displaystyle{ \frac{3(2x+3)}{3x^2+9x-5} }\)
Problem Statement |
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Calculate the derivative of this function and give your final answer in completely factored form. \(\displaystyle{ \ln(3x^2+9x-5) }\)
Solution |
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\(\displaystyle{ \frac{d}{dx}[ \ln(3x^2+9x-5) ] }\) |
Apply the chain rule. |
\(\displaystyle{ \frac{1}{3x^2+9x-5} \cdot \frac{d}{dx}[3x^2+9x-5] }\) |
\(\displaystyle{ \frac{1}{3x^2+9x-5} \cdot (6x+9) }\) |
\(\displaystyle{ \frac{3(2x+3)}{3x^2+9x-5} }\) |
Using the substitution \( u = 3x^2+9x-5 \) this could have solved more explicitly.
\(\displaystyle{ \frac{d}{dx}[ \ln(3x^2+9x-5) ] }\) |
\(\displaystyle{ \frac{d}{du}[\ln(u)] \cdot \frac{d}{dx}[3x^2+9x-5] }\) |
\(\displaystyle{ \frac{1}{u} \cdot (6x+9) }\) |
\(\displaystyle{ \frac{1}{3x^2+9x-5} \cdot [ 3(2x+3) ] }\) |
Final Answer |
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\(\displaystyle{ \frac{3(2x+3)}{3x^2+9x-5} }\) |
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\(h(x)=\ln(x^2+3x+4)\)
Problem Statement |
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Calculate the derivative of \(h(x)=\ln(x^2+3x+4)\)
Solution |
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1081 video
video by PatrickJMT |
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\(f(x)=\ln(5x^2+2x-7)\)
Problem Statement |
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Calculate the derivative of \(f(x)=\ln(5x^2+2x-7)\)
Solution |
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1156 video
video by MathTV |
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\(g(x)=\log_4(x^3+8x)\)
Problem Statement |
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Calculate the derivative of \(g(x)=\log_4(x^3+8x)\)
Solution |
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1082 video
video by PatrickJMT |
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\(\displaystyle{f(x)=\ln\sqrt[3]{x^3-x}}\)
Problem Statement |
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Calculate the derivative of \(\displaystyle{f(x)=\ln\sqrt[3]{x^3-x}}\)
Final Answer |
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\(\displaystyle{ \frac{3x^2-1}{3x(x^2-1)} }\)
Problem Statement |
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Calculate the derivative of \(\displaystyle{f(x)=\ln\sqrt[3]{x^3-x}}\)
Solution |
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1070 video
video by Krista King Math |
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Final Answer |
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\(\displaystyle{ \frac{3x^2-1}{3x(x^2-1)} }\) |
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\(\displaystyle{f(x)=\ln\left(x\sqrt{x^2+1}\right)}\)
Problem Statement |
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Calculate the derivative of \(\displaystyle{f(x)=\ln\left(x\sqrt{x^2+1}\right)}\)
Solution |
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1072 video
video by Krista King Math |
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Calculate the first three derivatives of \(\displaystyle{ f(x)=\ln(2+3x) }\) and give your answers in completely factored form.
Problem Statement |
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Calculate the first three derivatives of \(\displaystyle{ f(x)=\ln(2+3x) }\) and give your answers in completely factored form.
Solution |
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First Derivative
\(\displaystyle{ f'(x) = \frac{d}{dx}[ \ln(2+3x) ] }\) |
\(\displaystyle{ \frac{1}{2+3x} \frac{d}{dx}[2+3x] }\) |
\(\displaystyle{ \frac{1}{2+3x} (3) = \frac{3}{2+3x} }\) |
Second Derivative
We can use the either quotient rule or product rule here. The product rule is the easiest here because of the constant in the numerator but we need to rewrite the first derivative as \(\displaystyle{ \frac{3}{2+3x} = 3(2+3x)^{-1} }\).
\(\displaystyle{ f''(x) = \frac{d}{dx} \left[ 3(2+3x)^{-1} \right] }\) |
\(\displaystyle{3(-1)(2+3x)^{-2} \frac{d}{dx}[2+3x] }\) |
\(\displaystyle{-3(2+3x)^{-2} (3) = \frac{-9}{(2+3x)^{2}} }\) |
Third Derivative
Again, we can use either the quotient rule or product rule and we choose the product rule.
\( f^{(3)}(x) = \displaystyle{ \frac{d}{dx}\left[ -9(2+3x)^{-2} \right]} \) |
\(\displaystyle{-9(-2)(2+3x)^{-3}\frac{d}{dx}[2+3x]} \) |
\( 18(2+3x)^{-3} (3) \) |
\(\displaystyle{\frac{54}{(2+3x)^3} }\) |
Final Answers |
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\(\displaystyle{f'(x) = \frac{3}{2+3x}}\) |
\(\displaystyle{f''(x) = \frac{-9}{(2+3x)^{2}}}\) |
\(\displaystyle{ f^{(3)}(x)=\frac{54}{(2+3x)^3} }\) |
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Intermediate |
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\(\displaystyle{f(x)=\ln\left[\frac{(2x+1)^5}{\sqrt{x^2+1}}\right]}\)
Problem Statement |
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Calculate the derivative of \(\displaystyle{f(x)=\ln\left[\frac{(2x+1)^5}{\sqrt{x^2+1}}\right]}\)
Solution |
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1073 video
video by Krista King Math |
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\(\displaystyle{ f(x)=\ln\left[ \frac{(2x+1)^3}{(3x-1)^4} \right] }\)
Problem Statement |
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Calculate the derivative of \(\displaystyle{ f(x)=\ln\left[ \frac{(2x+1)^3}{(3x-1)^4} \right] }\)
Solution |
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1086 video
video by PatrickJMT |
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\(y=\sqrt[3]{\log_7(x)}\)
Problem Statement |
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Calculate the derivative of \(y=\sqrt[3]{\log_7(x)}\)
Solution |
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1089 video
video by PatrickJMT |
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\(y=\ln(x^4\sin x)\)
Problem Statement |
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Calculate the derivative of \(y=\ln(x^4\sin x)\)
Solution |
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1090 video
video by PatrickJMT |
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\(y=[\log_4(1+e^x)]^2\)
Problem Statement |
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Calculate the derivative of \(y=[\log_4(1+e^x)]^2\)
Solution |
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1087 video
video by PatrickJMT |
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\(p(x)=\ln\left[ x^2 \cdot \sqrt{x^3+3x} \cdot (x+2)^4 \right] \)
Problem Statement |
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Calculate the derivative of \(p(x)=\ln\left[ x^2 \cdot \sqrt{x^3+3x} \cdot (x+2)^4 \right] \)
Solution |
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1083 video
video by PatrickJMT |
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You CAN Ace Calculus
Topics You Need To Understand For This Page
For the basic logarithm derivatives you do not need the chain rule. But we discuss it on this page. Each section is labeled. So if you have not studied the chain rule yet, you can read the sections that apply to you and then come back here once you have studied it. |
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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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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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, calculate the derivative of these functions.
