Notes

Questions

Find the derivative

\(\textbf{1)}\) Find the derivative of \(f(x)=\ln x^3\)

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The answer is \(f'(x)=\displaystyle\frac{3}{x}\)

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\(\,\,\,\,\,f(x) = \ln(x^3)\)
\(\,\,\,\,\,f(x) = 3 \ln(x)\)
\(\,\,\,\,\,f'(x) = 3 \cdot \displaystyle \frac{1}{x}\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{3}{x}\)

 

\(\textbf{2)}\) Find the derivative of \(f(x)=(\ln⁡x)^6\)

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The answer is \(f'(x)=\displaystyle\frac{6(\ln⁡x)^5}{x}\)

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\(\,\,\,\,\,\text{This will use the chain rule}\)
\(\,\,\,\,\,f(x) = (\ln(x))^6\)
\(\,\,\,\,\,f'(x) = 6 (\ln(x))^5 \cdot \displaystyle \frac{1}{x}\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{6(\ln(x))^5}{x}\)

 

\(\textbf{3)}\) Find the derivative of \(f(x)=x \ln⁡x\)

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The answer is \(f'(x)=\ln⁡x+1\)

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\(\,\,\,\,\,\text{This will use the product rule}\)
\(\,\,\,\,\,f(x) = x \ln(x)\)
\(\,\,\,\,\,f'(x) = (x) \cdot \displaystyle \frac{1}{x} + (1) \cdot \ln(x)\)
\(\,\,\,\,\,f'(x) = \ln(x) + 1\)

 

\(\textbf{4)}\) Find the derivative of \(f(x)=\ln⁡(\ln⁡ x)\)

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The answer is \(f'(x)=\displaystyle\frac{1}{x \ln ⁡x }\)

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\(\,\,\,\,\,\text{This will use the chain rule}\)
\(\,\,\,\,\,f(x) = \ln(\ln(x))\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{1}{\ln(x)} \cdot \frac{1}{x}\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{1}{x \ln(x)}\)

 

\(\textbf{5)}\) Find the derivative of \(f(x)=\displaystyle\frac{\ln x⁡}{x}\)

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The answer is \(f'(x)=\displaystyle\frac{1-\ln ⁡ x}{x^2}\)

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\(\,\,\,\,\,\text{This will use the quotient rule}\)
\(\,\,\,\,\,f(x) = \displaystyle \frac{\ln(x)}{x}\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{\left(\frac{1}{x}\right) \cdot x – \ln(x) \cdot 1}{x^2}\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{1 – \ln(x)}{x^2}\)

 

\(\textbf{6)}\) Find the derivative of \(f(x)=\ln⁡(5x^2+2)\)

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The answer is \(f'(x)=\displaystyle\frac{10x}{5x^2+2}\)

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\(\,\,\,\,\,\text{This will use the chain rule}\)
\(\,\,\,\,\,f(x) = \ln(5x^2 + 2)\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{1}{5x^2 + 2} \cdot 10x\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{10x}{5x^2 + 2}\)

 

\(\textbf{7)}\) Find the derivative of \(f(x)=\ln⁡(\sqrt{x})\)

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The answer is \(f'(x)=\displaystyle\frac{1}{2x}\)

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\(\,\,\,\,\,\text{This will use the chain rule}\)
\(\,\,\,\,\,f(x) = \ln(\sqrt{x})\)
\(\,\,\,\,\,f(x) = \displaystyle \frac{1}{2} \ln(x)\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{1}{2} \cdot \frac{1}{x}\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{1}{2x}\)

 

\(\textbf{8)}\) Find the second derivative \(f”(x)\) of \(f(x)=\ln⁡{\left(3x^2\right)}\)

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The answer is \(f”(x)=\displaystyle -\frac{2}{x^2}\)

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\(\,\,\,\,\,\text{This will use the chain rule to find } f'(x) \text{, then find } f”(x)\)
\(\,\,\,\,\,f(x) = \ln(3x^2)\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{1}{3x^2} \cdot 6x\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{2}{x}\)
\(\,\,\,\,\,f”(x) = \displaystyle -\frac{2}{x^2}\)

 

\(\textbf{9)}\) Find the derivative of \(f(x)=x^2 \ln x\)

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The answer is \(f'(x)=2x \ln x + x\)

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\(\,\,\,\,\,\text{This will use the product rule}\)
\(\,\,\,\,\,f(x) = x^2 \ln(x)\)
\(\,\,\,\,\,f'(x) = (x^2) \cdot \displaystyle \frac{1}{x} + (2x) \cdot \ln(x)\)
\(\,\,\,\,\,f'(x) = 2x \ln(x) + x\)

 

\(\textbf{10)}\) Find the derivative of \(f(x)=\ln⁡(x^2+4)\)

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The answer is \(f'(x)=\displaystyle\frac{2x}{x^2+4}\)

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\(\,\,\,\,\,\text{This will use the chain rule}\)
\(\,\,\,\,\,f(x) = \ln(x^2 + 4)\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{1}{x^2 + 4} \cdot 2x\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{2x}{x^2 + 4}\)

 

\(\textbf{11)}\) Find the derivative of \(f(x)=\displaystyle \frac{\ln x}{x^3}\)

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The answer is \(f'(x)=\displaystyle \frac{1-3\ln x}{x^4}\)

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\(\,\,\,\,\,\text{This will use the quotient rule}\)
\(\,\,\,\,\,f(x) = \displaystyle \frac{\ln(x)}{x^3}\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{\left(\frac{1}{x}\right) \cdot x^3 – \ln(x) \cdot 3x^2}{x^6}\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{x^2 – 3x^2 \ln(x)}{x^6}\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{1 – 3\ln(x)}{x^4}\)

 

\(\textbf{12)}\) Find the derivative of \(f(x)=\ln(\sin x)\)

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The answer is \(f'(x)=\cot x\)

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\(\,\,\,\,\,\text{This will use the chain rule}\)
\(\,\,\,\,\,f(x) = \ln(\sin(x))\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{1}{\sin(x)} \cdot \cos(x)\)
\(\,\,\,\,\,f'(x) = \cot(x)\)

 

\(\textbf{13)}\) Find the derivative of \(f(x)=x \ln(x^2)\)

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The answer is \(f'(x)=\ln(x^2)+2\)

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\(\,\,\,\,\,\text{This will use the product rule and properties of logarithms}\)
\(\,\,\,\,\,f(x) = x \ln(x^2)\)
\(\,\,\,\,\,f(x) = 2x \ln(x)\)
\(\,\,\,\,\,f'(x) = 2 \ln(x) + x \cdot \displaystyle \frac{2}{x}\)
\(\,\,\,\,\,f'(x) = \ln(x^2) + 2\)

 

\(\textbf{14)}\) Find the derivative of \(f(x)=\ln(x^3+1)\)

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The answer is \(f'(x)=\displaystyle\frac{3x^2}{x^3+1}\)

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\(\,\,\,\,\,\text{This will use the chain rule}\)
\(\,\,\,\,\,f(x) = \ln(x^3 + 1)\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{1}{x^3 + 1} \cdot 3x^2\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{3x^2}{x^3 + 1}\)

 

 

See Related Pages\(\)

\(\bullet\text{ Calculus Homepage}\)
\(\,\,\,\,\,\,\,\,\text{All the Best Topics…}\)

\(\bullet\text{ Definition of Derivative}\)
\(\,\,\,\,\,\,\,\, \displaystyle \lim_{\Delta x\to 0} \frac{f(x+ \Delta x)-f(x)}{\Delta x} \)

\(\bullet\text{ Equation of the Tangent Line}\)
\(\,\,\,\,\,\,\,\,f(x)=x^3+3x^2−x \text{ at the point } (2,18)\)

\(\bullet\text{ Derivatives- Constant Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(c)=0\)

\(\bullet\text{ Derivatives- Power Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(x^n)=nx^{n-1}\)

\(\bullet\text{ Derivatives- Constant Multiple Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(cf(x))=cf'(x)\)

\(\bullet\text{ Derivatives- Sum and Difference Rules}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(x) \pm g(x)]=f'(x) \pm g'(x)\)

\(\bullet\text{ Derivatives- Sin and Cos}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}sin(x)=cos(x)\)

\(\bullet\text{ Derivatives- Product Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(x) \cdot g(x)]=f(x) \cdot g'(x)+f'(x) \cdot g(x)\)

\(\bullet\text{ Derivatives- Quotient Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\displaystyle\frac{f(x)}{g(x)}\right]=\displaystyle\frac{g(x) \cdot f'(x)-f(x) \cdot g'(x)}{[g(x)]^2}\)

\(\bullet\text{ Derivatives- Chain Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(g(x))]= f'(g(x)) \cdot g'(x)\)

\(\bullet\text{ Derivatives- ln(x)}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[ln(x)]= \displaystyle \frac{1}{x}\)

\(\bullet\text{ Implicit Differentiation}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Horizontal Tangent Line}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Mean Value Theorem}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Related Rates}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Increasing and Decreasing Intervals}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Intervals of concave up and down}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Inflection Points}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Graph of f(x), f'(x) and f”(x)}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Newton’s Method}\)
\(\,\,\,\,\,\,\,\,x_{n+1}=x_n – \displaystyle \frac{f(x_n)}{f'(x_n)}\)

 

In Summary

Logarithmic functions are the inverse operation of exponential functions. The derivatives of logarithmic functions are very interesting.

\(\frac{d}{dx} \ln x = \frac{1}{x}\) and \(\frac{d}{dx} \log_a x = \frac{1}{x \ln a}\).

The derivative of ln(x) is usually introduced in a calculus course while learning about other derivatives.