The derivative of the natural logarithm is one of the most important derivative rules in calculus. Since \(\frac{d}{dx}\ln(x)=\frac{1}{x}\), logarithmic derivatives often combine with the chain rule, product rule, and quotient rule. These problems focus on finding derivatives of natural log expressions, including nested logs, products, quotients, and second derivatives.
Notes
Practice Problems
Find the derivative
\(\textbf{1)}\) Find the derivative of \(f(x)=\ln x^3\)
The answer is \(f'(x)=\displaystyle\frac{3}{x}\)
\(\,\,\,\,\,f(x) = \ln(x^3)\)
\(\,\,\,\,\,f(x) = 3 \ln(x)\)
\(\,\,\,\,\,f'(x) = 3 \cdot \displaystyle \frac{1}{x}\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{3}{x}\)
\(\,\,\,\,\,\)The answer is \(f'(x)=\displaystyle\frac{3}{x}\)
\(\textbf{2)}\) Find the derivative of \(f(x)=(\lnx)^6\)
The answer is \(f'(x)=\displaystyle\frac{6(\lnx)^5}{x}\)
\(\,\,\,\,\,\text{This will use the chain rule}\)
\(\,\,\,\,\,f(x) = (\ln(x))^6\)
\(\,\,\,\,\,f'(x) = 6 (\ln(x))^5 \cdot \displaystyle \frac{1}{x}\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{6(\ln(x))^5}{x}\)
\(\,\,\,\,\,\)The answer is \(f'(x)=\displaystyle\frac{6(\lnx)^5}{x}\)
\(\textbf{3)}\) Find the derivative of \(f(x)=x \lnx\)
The answer is \(f'(x)=\lnx+1\)
\(\,\,\,\,\,\text{This will use the product rule}\)
\(\,\,\,\,\,f(x) = x \ln(x)\)
\(\,\,\,\,\,f'(x) = (x) \cdot \displaystyle \frac{1}{x} + (1) \cdot \ln(x)\)
\(\,\,\,\,\,f'(x) = \ln(x) + 1\)
\(\,\,\,\,\,\)The answer is \(f'(x)=\lnx+1\)
\(\textbf{4)}\) Find the derivative of \(f(x)=\ln(\ln x)\)
The answer is \(f'(x)=\displaystyle\frac{1}{x \ln x }\)
\(\,\,\,\,\,\text{This will use the chain rule}\)
\(\,\,\,\,\,f(x) = \ln(\ln(x))\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{1}{\ln(x)} \cdot \frac{1}{x}\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{1}{x \ln(x)}\)
\(\,\,\,\,\,\)The answer is \(f'(x)=\displaystyle\frac{1}{x \ln x }\)
\(\textbf{5)}\) Find the derivative of \(f(x)=\displaystyle\frac{\ln x}{x}\)
The answer is \(f'(x)=\displaystyle\frac{1-\ln x}{x^2}\)
\(\,\,\,\,\,\text{This will use the quotient rule}\)
\(\,\,\,\,\,f(x) = \displaystyle \frac{\ln(x)}{x}\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{\left(\frac{1}{x}\right) \cdot x – \ln(x) \cdot 1}{x^2}\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{1 – \ln(x)}{x^2}\)
\(\,\,\,\,\,\)The answer is \(f'(x)=\displaystyle\frac{1-\ln x}{x^2}\)
\(\textbf{6)}\) Find the derivative of \(f(x)=\ln(5x^2+2)\)
The answer is \(f'(x)=\displaystyle\frac{10x}{5x^2+2}\)
\(\,\,\,\,\,\text{This will use the chain rule}\)
\(\,\,\,\,\,f(x) = \ln(5x^2 + 2)\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{1}{5x^2 + 2} \cdot 10x\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{10x}{5x^2 + 2}\)
\(\,\,\,\,\,\)The answer is \(f'(x)=\displaystyle\frac{10x}{5x^2+2}\)
\(\textbf{7)}\) Find the derivative of \(f(x)=\ln(\sqrt{x})\)
The answer is \(f'(x)=\displaystyle\frac{1}{2x}\)
\(\,\,\,\,\,\text{This will use the chain rule}\)
\(\,\,\,\,\,f(x) = \ln(\sqrt{x})\)
\(\,\,\,\,\,f(x) = \displaystyle \frac{1}{2} \ln(x)\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{1}{2} \cdot \frac{1}{x}\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{1}{2x}\)
\(\,\,\,\,\,\)The answer is \(f'(x)=\displaystyle\frac{1}{2x}\)
\(\textbf{8)}\) Find the second derivative \(f”(x)\) of \(f(x)=\ln{\left(3x^2\right)}\)
The answer is \(f”(x)=\displaystyle -\frac{2}{x^2}\)
\(\,\,\,\,\,\text{This will use the chain rule to find } f'(x) \text{, then find } f”(x)\)
\(\,\,\,\,\,f(x) = \ln(3x^2)\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{1}{3x^2} \cdot 6x\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{2}{x}\)
\(\,\,\,\,\,f”(x) = \displaystyle -\frac{2}{x^2}\)
\(\,\,\,\,\,\)The answer is \(f”(x)=\displaystyle -\frac{2}{x^2}\)
\(\textbf{9)}\) Find the derivative of \(f(x)=x^2 \ln x\)
The answer is \(f'(x)=2x \ln x + x\)
\(\,\,\,\,\,\text{This will use the product rule}\)
\(\,\,\,\,\,f(x) = x^2 \ln(x)\)
\(\,\,\,\,\,f'(x) = (x^2) \cdot \displaystyle \frac{1}{x} + (2x) \cdot \ln(x)\)
\(\,\,\,\,\,f'(x) = 2x \ln(x) + x\)
\(\,\,\,\,\,\)The answer is \(f'(x)=2x \ln x + x\)
\(\textbf{10)}\) Find the derivative of \(f(x)=\ln(x^2+4)\)
The answer is \(f'(x)=\displaystyle\frac{2x}{x^2+4}\)
\(\,\,\,\,\,\text{This will use the chain rule}\)
\(\,\,\,\,\,f(x) = \ln(x^2 + 4)\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{1}{x^2 + 4} \cdot 2x\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{2x}{x^2 + 4}\)
\(\,\,\,\,\,\)The answer is \(f'(x)=\displaystyle\frac{2x}{x^2+4}\)
\(\textbf{11)}\) Find the derivative of \(f(x)=\displaystyle \frac{\ln x}{x^3}\)
The answer is \(f'(x)=\displaystyle \frac{1-3\ln x}{x^4}\)
\(\,\,\,\,\,\text{This will use the quotient rule}\)
\(\,\,\,\,\,f(x) = \displaystyle \frac{\ln(x)}{x^3}\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{\left(\frac{1}{x}\right) \cdot x^3 – \ln(x) \cdot 3x^2}{x^6}\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{x^2 – 3x^2 \ln(x)}{x^6}\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{1 – 3\ln(x)}{x^4}\)
\(\,\,\,\,\,\)The answer is \(f'(x)=\displaystyle \frac{1-3\ln x}{x^4}\)
\(\textbf{12)}\) Find the derivative of \(f(x)=\ln(\sin x)\)
The answer is \(f'(x)=\cot x\)
\(\,\,\,\,\,\text{This will use the chain rule}\)
\(\,\,\,\,\,f(x) = \ln(\sin(x))\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{1}{\sin(x)} \cdot \cos(x)\)
\(\,\,\,\,\,f'(x) = \cot(x)\)
\(\,\,\,\,\,\)The answer is \(f'(x)=\cot x\)
\(\textbf{13)}\) Find the derivative of \(f(x)=x \ln(x^2)\)
The answer is \(f'(x)=\ln(x^2)+2\)
\(\,\,\,\,\,\text{This will use the product rule and properties of logarithms}\)
\(\,\,\,\,\,f(x) = x \ln(x^2)\)
\(\,\,\,\,\,f'(x)=1\cdot\ln(x^2)+x\cdot\frac{2x}{x^2}\)
\(\,\,\,\,\,f'(x)=\ln(x^2)+x\cdot\frac{2}{x}\)
\(\,\,\,\,\,f'(x) = \ln(x^2) + 2\)
\(\,\,\,\,\,\)The answer is \(f'(x)=\ln(x^2)+2\)
\(\textbf{14)}\) Find the derivative of \(f(x)=\ln(x^3+1)\)
The answer is \(f'(x)=\displaystyle\frac{3x^2}{x^3+1}\)
\(\,\,\,\,\,\text{This will use the chain rule}\)
\(\,\,\,\,\,f(x) = \ln(x^3 + 1)\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{1}{x^3 + 1} \cdot 3x^2\)
\(\,\,\,\,\,f'(x) = \displaystyle \frac{3x^2}{x^3 + 1}\)
\(\,\,\,\,\,\)The answer is \(f'(x)=\displaystyle\frac{3x^2}{x^3+1}\)
\(\textbf{15)}\) Find the derivative of \(f(x)=\ln(7x-4)\)
The answer is \(f'(x)=\displaystyle\frac{7}{7x-4}\)
\(\,\,\,\,\,\text{This will use the chain rule}\)
\(\,\,\,\,\,f(x)=\ln(7x-4)\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{1}{7x-4}\cdot 7\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{7}{7x-4}\)
\(\,\,\,\,\,\)The answer is \(f'(x)=\displaystyle\frac{7}{7x-4}\)
\(\textbf{16)}\) Find the derivative of \(f(x)=\ln(x^2+6x+10)\)
The answer is \(f'(x)=\displaystyle\frac{2x+6}{x^2+6x+10}\)
\(\,\,\,\,\,\text{This will use the chain rule}\)
\(\,\,\,\,\,f(x)=\ln(x^2+6x+10)\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{1}{x^2+6x+10}\cdot(2x+6)\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{2x+6}{x^2+6x+10}\)
\(\,\,\,\,\,\)The answer is \(f'(x)=\displaystyle\frac{2x+6}{x^2+6x+10}\)
\(\textbf{17)}\) Find the derivative of \(f(x)=\ln\left(\frac{x+1}{x-1}\right)\)
The answer is \(f'(x)=\displaystyle-\frac{2}{x^2-1}\)
\(\,\,\,\,\,f(x)=\ln\left(\frac{x+1}{x-1}\right)\)
\(\,\,\,\,\,f(x)=\ln(x+1)-\ln(x-1)\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{1}{x+1}-\frac{1}{x-1}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{x-1-(x+1)}{(x+1)(x-1)}\)
\(\,\,\,\,\,f'(x)=\displaystyle-\frac{2}{x^2-1}\)
\(\,\,\,\,\,\)The answer is \(f'(x)=\displaystyle-\frac{2}{x^2-1}\)
\(\textbf{18)}\) Find the derivative of \(f(x)=\ln\left(e^x+x\right)\)
The answer is \(f'(x)=\displaystyle\frac{e^x+1}{e^x+x}\)
\(\,\,\,\,\,\text{This will use the chain rule}\)
\(\,\,\,\,\,f(x)=\ln\left(e^x+x\right)\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{1}{e^x+x}\cdot(e^x+1)\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{e^x+1}{e^x+x}\)
\(\,\,\,\,\,\)The answer is \(f'(x)=\displaystyle\frac{e^x+1}{e^x+x}\)
\(\textbf{19)}\) Find the derivative of \(f(x)=\ln(\cos x)\)
The answer is \(f'(x)=-\tan x\)
\(\,\,\,\,\,\text{This will use the chain rule}\)
\(\,\,\,\,\,f(x)=\ln(\cos x)\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{1}{\cos x}\cdot(-\sin x)\)
\(\,\,\,\,\,f'(x)=-\displaystyle\frac{\sin x}{\cos x}\)
\(\,\,\,\,\,f'(x)=-\tan x\)
\(\,\,\,\,\,\)The answer is \(f'(x)=-\tan x\)
\(\textbf{20)}\) Find the second derivative \(f”(x)\) of \(f(x)=\ln(x^2+1)\)
The answer is \(f”(x)=\displaystyle\frac{2-2x^2}{(x^2+1)^2}\)
\(\,\,\,\,\,f(x)=\ln(x^2+1)\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{2x}{x^2+1}\)
\(\,\,\,\,\,\text{Use the quotient rule to find }f”(x).\)
\(\,\,\,\,\,f”(x)=\displaystyle\frac{(x^2+1)(2)-2x(2x)}{(x^2+1)^2}\)
\(\,\,\,\,\,f”(x)=\displaystyle\frac{2x^2+2-4x^2}{(x^2+1)^2}\)
\(\,\,\,\,\,f”(x)=\displaystyle\frac{2-2x^2}{(x^2+1)^2}\)
\(\,\,\,\,\,\)The answer is \(f”(x)=\displaystyle\frac{2-2x^2}{(x^2+1)^2}\)
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