Derivative of ln(x)

\[\bbox[yellow,5px]{\dfrac{d}{dx}\ln x = \dfrac{1}{x} }\]
Applying the Chain rule, we have
\[\dfrac{d}{dx}\ln(\text{stuff}) = \dfrac{1}{(\text{stuff})} \cdot \dfrac{d}{dx} (\text{stuff}) \]
You’ll usually see this written as
\[\dfrac{d}{dx}\ln u = \dfrac{1}{u} \cdot \dfrac{du}{dx} \]

Problem #1

Differentiate $f(x) = \ln (x^2 + 3x -1).$

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\begin{align*} f'(x) &= \frac{d}{dx} \left[ \ln (x^2 + 3x -1) \right] \\[8px] &= \frac{1}{(x^2 + 3x -1)} \cdot \left[ \frac{d}{dx}(x^2 + 3x -1) \right] \\[8px] &= \frac{2x + 3}{(x^2 + 3x -1)} \quad \cmark \end{align*}

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Problem #2

Differentiate $f(x) = \sqrt{\ln x^2}.$

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\begin{align*} f'(x) &= \frac{d}{dx} \left[ \sqrt{\ln x^2}\right]\\[8px] &= \dfrac{d}{dx} \left[ \ln x^2\right]^{1/2} \\[8px] &=\frac{1}{2} \left[ \ln x^2\right]^{-1/2} \cdot \left[ \frac{d}{dx}\ln x^2\right] \\[8px] &=\frac{1}{2} \frac{1}{\sqrt{\ln x^2}} \cdot \left[ \frac{1}{x^2} \left(\frac{d}{dx} x^2 \right)\right] \\[8px] &= \frac{1}{2} \frac{1}{\sqrt{\ln x^2}} \cdot \frac{1}{x^2} (2x) \\[8px] &= \frac{1}{x \sqrt{\ln x^2}} \quad \cmark \end{align*}

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Problem #3

Differentiate $f(x) = \ln(\ln x).$

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\begin{align*} f'(x) &= \frac{d}{dx} \left[ \ln(\ln x)\right] \\[8px] &= \frac{1}{\ln x} \cdot \frac{d}{dx}\left( \ln x \right) \\[8px] &= \frac{1}{\ln x} \cdot \frac{1}{x} \\[8px] &= \frac{1}{x \ln x} \quad \cmark \end{align*}

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Problem #4

Differentiate $f(x) = \ln(x^2 e^x).$

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We’ll solve this two ways:
Method 1: Chain rule and Product rule
\begin{align*} \dfrac{d}{dx}\ln(x^2 e^x) &= \dfrac{1}{x^2 e^x} \cdot \dfrac{d}{dx} \left[ x^2 e^x\right] \\[8px] &= \dfrac{1}{x^2 e^x} \cdot \left[\left(\dfrac{d}{dx}x^2 \right)e^x + x^2 \left( \dfrac{d}{dx} e^x\right) \right] \\[8px] &= \dfrac{1}{x^2 e^x} \cdot \left[2x e^x + x^2 e^x \right] \\[8px] &= \dfrac{2x e^x}{x^2 e^x} + \dfrac{x^2 e^x}{x^2 e^x} \\[8px] &= \dfrac{2}{x} + 1 \quad \cmark \end{align*} Method 2: Use log property.
Recall that \begin{align*} \ln(f*g) &= \ln(f) + \ln(g) \\[8px] \text{Hence} \phantom{\ln(f*g)}& \\[8px] \ln(x^2 e^x) &= \ln(x^2) + \ln(e^x) \end{align*} Then we can compute the derivative: \begin{align*} \dfrac{d}{dx}\ln(x^2 e^x) &= \dfrac{d}{dx}\ln(x^2) + \dfrac{d}{dx}\ln(e^x) \\[8px] &= \dfrac{1}{x^2} \cdot \left(\dfrac{d}{dx}x^2 \right) + \dfrac{1}{e^x} \cdot \left(\dfrac{d}{dx}e^x \right) \\[8px] &= \dfrac{1}{x^2} \cdot 2x + \dfrac{1}{e^x} \cdot e^x \\[8px] &= \dfrac{2}{x} + 1 \quad \cmark \end{align*}

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Problem #5

Differentiate $f(x) = \ln\left(\dfrac{\cos x}{5x} \right).$

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We’ll solve this two ways.
Method 1: Chain rule and Quotient rule.
\begin{align*} \dfrac{d}{dx}\left[\ln\left(\dfrac{\cos x}{5x} \right) \right] &= \dfrac{1}{\frac{\cos x}{5x}} \cdot \dfrac{d}{dx}\left[\dfrac{\cos x}{5x} \right] \\[8px] &= \dfrac{5x}{\cos x} \cdot \left[\dfrac{\left(\dfrac{d}{dx}\cos x \right)5x – \cos x \left(\dfrac{d}{dx}5x \right) }{(5x)^2} \right] \\[8px] &= \dfrac{5x}{\cos x} \cdot \left[\dfrac{(-\sin x)5x – \cos x (5)}{(5x)^2} \right] \\[8px] &= \dfrac{1}{\cos x} \left[\dfrac{-5x \sin x – 5 \cos x}{5x} \right] \\[8px] &= \dfrac{-5x\sin x}{5x\cos x} – \dfrac{5 \cos x}{5x \cos x} \\[8px] &= -\tan x – \dfrac{1}{x} \quad \cmark \end{align*} Method 2: Use log property
Recall that Recall that \begin{align*} \ln\left(\dfrac{f}{g} \right) &= \ln(f) – \ln(g) \\[8px] \text{Hence} \phantom{\ln(f*g)}& \\[8px] \ln\left(\dfrac{\cos x}{5x} \right) &= \ln(\cos x) – \ln(5x) \end{align*} Then we can compute the derivative: \begin{align*} \dfrac{d}{dx}\left[\ln\left(\dfrac{\cos x}{5x} \right) \right] &= \dfrac{d}{dx}\ln(\cos x) – \dfrac{d}{dx}\ln(5x) \\[8px] &= \dfrac{1}{\cos x} \cdot \left(\dfrac{d}{dx}\cos x \right) – \dfrac{1}{5x} \cdot \left(\dfrac{d}{dx}5x \right) \\[8px] &= \dfrac{1}{\cos x} \cdot (-\sin x) – \dfrac{1}{5x} \cdot (5) \\[8px] &= -\tan x – \dfrac{1}{x} \quad \cmark \end{align*}

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