Logarithmic Differentiation
Basic Rule of Logarithmic Differentiation
$$ \displaystyle \dfrac{d}{dx}\log_e{x} = \dfrac{1}{x} \\
\dfrac{d}{dx}\log_e{f(x)} = \dfrac{f'(x)}{f(x)} $$
Practice Questions
Question 1
Differentiate \( y = \log_{e}(3x) \).
\( \begin{aligned} \displaystyle
\dfrac{d}{dx}\log_{e}(3x) &= \dfrac{(3x)’}{3x} \\
&= \dfrac{3}{3x} \\
&= \dfrac{1}{x}
\end{aligned} \)
Question 2
Differentiate \( y = \log_{e}(2x-1) \).
\( \begin{aligned} \displaystyle
\dfrac{d}{dx}\log_{e}(2x-1) &= \dfrac{(2x-1)’}{2x-1} \\
&= \dfrac{2}{2x-1}
\end{aligned} \)
Question 3
Differentiate \( x^2\log_{e}x \).
\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\dfrac{d}{dx}x^2\log_{e}x &= 2x\log_{e}x + x^2 \times \dfrac{1}{x} \ \ \ \ \color{red} \text{product rule} \\
&= 2x\log_{e}x + x \\
\end{aligned} \)
Question 4
Differentiate \( \log_{e}\dfrac{x}{x-1} \).
\( \begin{aligned} \displaystyle
\log_{e}\dfrac{x}{x-1} &= \log_{e}x-\log_{e}(x-1) \\
\dfrac{d}{dx}\log_{e}\dfrac{x}{x-1} &= \dfrac{1}{x}-\dfrac{1}{x-1}
\end{aligned} \)
Responses