## 17Calculus Derivatives - Logarithm Functions

The logarithm rule is not as simple as the exponential derivative but it is still very straightforward.

Basic Logarithm Rule

$$\displaystyle{ \frac{d}{dt}[\ln(t)] = \frac{1}{t} }$$

Logarithm With Chain Rule

$$\displaystyle{ \frac{d}{dt}[\ln(u)] = \frac{1}{u}\frac{du}{dt} }$$

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

Many times, it helps to simplify a logarithmic expression before taking the derivative. Here are a few rules that should help you.

 $$\ln(xy) = \ln(x) + \ln(y)$$ $$\ln(x/y) = \ln(x) - \ln(y)$$ $$\ln(x^y) = y \ln(x)$$ review logarithms

Here is a short video clip that goes through these equations again.

### PatrickJMT - Derivatives of Logarithmic Functions and Examples [1min-29secs]

video by PatrickJMT

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

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

Calculate the derivative of $$\displaystyle{f(x)=\frac{1}{\ln(x)}}$$

Solution

### 1071 video

video by Krista King Math

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$$f(x)=\sqrt{x}\ln(x)$$

Problem Statement

Calculate the derivative of $$f(x)=\sqrt{x}\ln(x)$$

Solution

### 1085 video

video by PatrickJMT

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$$\displaystyle{y=\frac{\ln x}{1+\ln(2x)}}$$

Problem Statement

Calculate the derivative of $$\displaystyle{y=\frac{\ln x}{1+\ln(2x)}}$$

Solution

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

Calculate the derivative of $$y=\ln(x)+2^x+\sin(x)$$

Solution

### 1080 video

video by PatrickJMT

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These practice problems require the chain rule.

Basic

$$f(x)=\ln(x^2+10)$$

Problem Statement

Calculate the derivative of $$f(x)=\ln(x^2+10)$$

Solution

### 1084 video

video by PatrickJMT

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$$y=\ln(x^2+x)$$

Problem Statement

Calculate the derivative of $$y=\ln(x^2+x)$$

Solution

### 1088 video

video by PatrickJMT

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$$\displaystyle{ \ln(3x^2+9x-5) }$$

Problem Statement

Calculate the derivative of this function and give your final answer in completely factored form. $$\displaystyle{ \ln(3x^2+9x-5) }$$

$$\displaystyle{ \frac{3(2x+3)}{3x^2+9x-5} }$$

Problem Statement

Calculate the derivative of this function and give your final answer in completely factored form. $$\displaystyle{ \ln(3x^2+9x-5) }$$

Solution

 $$\displaystyle{ \frac{d}{dx}[ \ln(3x^2+9x-5) ] }$$ 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.

 $$\displaystyle{ \frac{d}{dx}[ \ln(3x^2+9x-5) ] }$$ $$\displaystyle{ \frac{d}{du}[\ln(u)] \cdot \frac{d}{dx}[3x^2+9x-5] }$$ $$\displaystyle{ \frac{1}{u} \cdot (6x+9) }$$ $$\displaystyle{ \frac{1}{3x^2+9x-5} \cdot [ 3(2x+3) ] }$$

$$\displaystyle{ \frac{3(2x+3)}{3x^2+9x-5} }$$

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$$h(x)=\ln(x^2+3x+4)$$

Problem Statement

Calculate the derivative of $$h(x)=\ln(x^2+3x+4)$$

Solution

### 1081 video

video by PatrickJMT

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$$f(x)=\ln(5x^2+2x-7)$$

Problem Statement

Calculate the derivative of $$f(x)=\ln(5x^2+2x-7)$$

Solution

### 1156 video

video by MathTV

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$$g(x)=\log_4(x^3+8x)$$

Problem Statement

Calculate the derivative of $$g(x)=\log_4(x^3+8x)$$

Solution

### 1082 video

video by PatrickJMT

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$$\displaystyle{f(x)=\ln\sqrt{x^3-x}}$$

Problem Statement

Calculate the derivative of $$\displaystyle{f(x)=\ln\sqrt{x^3-x}}$$

$$\displaystyle{ \frac{3x^2-1}{3x(x^2-1)} }$$

Problem Statement

Calculate the derivative of $$\displaystyle{f(x)=\ln\sqrt{x^3-x}}$$

Solution

### 1070 video

video by Krista King Math

$$\displaystyle{ \frac{3x^2-1}{3x(x^2-1)} }$$

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$$\displaystyle{f(x)=\ln\left(x\sqrt{x^2+1}\right)}$$

Problem Statement

Calculate the derivative of $$\displaystyle{f(x)=\ln\left(x\sqrt{x^2+1}\right)}$$

Solution

### 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

Calculate the first three derivatives of $$\displaystyle{ f(x)=\ln(2+3x) }$$ and give your answers in completely factored form.

Solution

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.

 $$f^{(3)}(x) = \displaystyle{ \frac{d}{dx}\left[ -9(2+3x)^{-2} \right]}$$ $$\displaystyle{-9(-2)(2+3x)^{-3}\frac{d}{dx}[2+3x]}$$ $$18(2+3x)^{-3} (3)$$ $$\displaystyle{\frac{54}{(2+3x)^3} }$$

$$\displaystyle{f'(x) = \frac{3}{2+3x}}$$

$$\displaystyle{f''(x) = \frac{-9}{(2+3x)^{2}}}$$

$$\displaystyle{ f^{(3)}(x)=\frac{54}{(2+3x)^3} }$$

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Intermediate

$$\displaystyle{f(x)=\ln\left[\frac{(2x+1)^5}{\sqrt{x^2+1}}\right]}$$

Problem Statement

Calculate the derivative of $$\displaystyle{f(x)=\ln\left[\frac{(2x+1)^5}{\sqrt{x^2+1}}\right]}$$

Solution

### 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

Calculate the derivative of $$\displaystyle{ f(x)=\ln\left[ \frac{(2x+1)^3}{(3x-1)^4} \right] }$$

Solution

### 1086 video

video by PatrickJMT

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$$y=\sqrt{\log_7(x)}$$

Problem Statement

Calculate the derivative of $$y=\sqrt{\log_7(x)}$$

Solution

### 1089 video

video by PatrickJMT

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$$y=\ln(x^4\sin x)$$

Problem Statement

Calculate the derivative of $$y=\ln(x^4\sin x)$$

Solution

### 1090 video

video by PatrickJMT

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$$y=[\log_4(1+e^x)]^2$$

Problem Statement

Calculate the derivative of $$y=[\log_4(1+e^x)]^2$$

Solution

### 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

Calculate the derivative of $$p(x)=\ln\left[ x^2 \cdot \sqrt{x^3+3x} \cdot (x+2)^4 \right]$$

Solution

### 1083 video

video by PatrickJMT

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You CAN Ace Calculus

 basic derivative rules power rule product rule quotient rule 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.

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia] Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

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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