# Calculus: Derivatives I

## Logarithmic Functions

The derivative of the natural logarithm is given by the equation below:

$$\dfrac{d}{dx} \ln(x) = \dfrac{1}{x}$$

You may have observed that the derivative of $x^0$ is $0$ and noticed this left a gap in the chain of differentiating polynomials. After all, $x^2$ becomes $2x$, $x$ becomes $1$, but $1$ becomes 0, although the pattern picks back up again as $x^{-1}$ becomes $-x^{-2}$. Thinking backwards, it seems like $x^{-1}$ must be the derivative of something, right? Well, it is. It's the derivative of the natural logarithm. Weird? Maybe. True? All day everyday.

## Problems

1. Differentiate with respect to $x$: $y=\ln(x) + 2$

$\dfrac{d}{dx}y = \dfrac{d}{dx}\left(\ln(x) + 2 \right) \\ \dfrac{dy}{dx} = \dfrac{d}{dx}\ln(x) + \dfrac{d}{dx}2 \\ \dfrac{dy}{dx} = \dfrac{1}{x} + 0 \\ \dfrac{dy}{dx} = \dfrac{1}{x}\\$

2. Differentiate with respect to $x$: $y=\ln(x) + \log_{2}(x)$

$\dfrac{d}{dx}y = \dfrac{d}{dx}\left(\ln(x) + \log_{2}(x) \right) \\ \dfrac{dy}{dx} = \dfrac{d}{dx}\ln(x) + \dfrac{d}{dx}\log_{2}(x) \\ \dfrac{dy}{dx} = \dfrac{d}{dx}\ln(x) + \dfrac{d}{dx}\dfrac{\ln(x)}{\ln(2)} \\ \dfrac{dy}{dx} = \dfrac{d}{dx}\ln(x) + \dfrac{1}{\ln(2)}\dfrac{d}{dx}\ln(x)\\ \dfrac{dy}{dx} = \dfrac{1}{x} + \dfrac{1}{\ln(2)} \cdot \dfrac{1}{x} \\ \dfrac{dy}{dx} = \dfrac{1}{x} + \dfrac{1}{\ln(2)x} \\$
3. Differentiate with respect to $x$: $y=-2\ln(x)$

$\dfrac{d}{dx}y = \dfrac{d}{dx}\left(-2\ln(x)\right) \\ \dfrac{dy}{dx} = -2\dfrac{d}{dx}\ln(x) \\ \dfrac{d}{dx} = -2\cdot\dfrac{1}{x}\\ \dfrac{d}{dx} = \dfrac{-2}{x}\\$
4. Differentiate with respect to $x$: $y=-2\log_{2}(x)$

$\dfrac{d}{dx}y = \dfrac{d}{dx}\left(-2\log_{2}(x)\right) \\ \dfrac{dy}{dx} = -2\dfrac{d}{dx}\log_{2}(x) \\ \dfrac{dy}{dx} = -2\dfrac{d}{dx}\dfrac{\ln(x)}{\ln(2)} \\ \dfrac{dy}{dx} = \dfrac{-2}{\ln(2)}\dfrac{d}{dx}\ln(x) \\ \dfrac{dy}{dx} = \dfrac{-2}{\ln(2)}\cdot \dfrac{1}{x} \\ \dfrac{dy}{dx} = \dfrac{-2}{\ln(2)x} \\$

5. Differentiate with respect to $x$: $y=\ln(2x) + \ln(x^2)$

$\dfrac{d}{dx}y = \dfrac{d}{dx}\left(\ln(2x) + \ln(x^2)\right) \\ \dfrac{dy}{dx} = \dfrac{d}{dx}\left(\ln(2) + \ln(x) + 2\ln(x) \right) \\ \dfrac{dy}{dx} = \dfrac{d}{dx}\ln(2) + \dfrac{d}{dx}\ln(x) + 2\dfrac{d}{dx}\ln(x) \\ \dfrac{dy}{dx} = 0 + \dfrac{1}{x} + 2\cdot\dfrac{1}{x} \\ \dfrac{dy}{dx} = \dfrac{3}{x} \\$
6. Differentiate with respect to $x$: $y=\ln(3x^3)$

$\dfrac{d}{dx}y = \dfrac{d}{dx}\ln(3x^3) \\ \dfrac{dy}{dx} = \dfrac{d}{dx}\left(\ln(3) + \ln(x^3) \right) \\ \dfrac{dy}{dx} = \dfrac{d}{dx}\left(\ln(3) + 3\ln(x) \right) \\ \dfrac{dy}{dx} = \dfrac{d}{dx}\ln(3) + 3\dfrac{d}{dx}\ln(x) \\ \dfrac{dy}{dx} = 0 + 3\cdot\dfrac{1}{x} \\ \dfrac{dy}{dx} = \dfrac{3}{x} \\$
7. Differentiate with respect to $x$: $y=\log_{3}(3x)$

$\dfrac{d}{dx}y = \dfrac{d}{dx}\log_{3}(3x) \\ \dfrac{dy}{dx} = \dfrac{d}{dx}\left( \log_{3}(3) + \log_{3}(x) \right) \\ \dfrac{dy}{dx} = \dfrac{d}{dx} \log_{3}(3) + \dfrac{d}{dx}\log_{3}(x) \\ \dfrac{dy}{dx} = \dfrac{d}{dx} \log_{3}(3) + \dfrac{d}{dx}\dfrac{\ln(x)}{\ln(3)} \\ \dfrac{dy}{dx} = \dfrac{d}{dx} \log_{3}(3) + \dfrac{1}{\ln(3)}\dfrac{d}{dx}\ln(x) \\ \dfrac{dy}{dx} = 0 + \dfrac{1}{\ln(3)} \cdot \dfrac{1}{x} \\ \dfrac{dy}{dx} = \dfrac{1}{\ln(3)x} \\$
8. Differentiate with respect to $x$: $y=\dfrac{\log_{8}(2x)}{\ln(4)}$

$\dfrac{d}{dx}y = \dfrac{d}{dx}\dfrac{\log_{8}(2x)}{\ln(4)} \\ \dfrac{dy}{dx} = \dfrac{d}{dx}\dfrac{\ln(2x)}{\ln(8)\ln(4)} \\ \dfrac{dy}{dx} = \dfrac{d}{dx}\dfrac{\ln(2) + \ln(x)}{\ln(8)\ln(4)} \\ \dfrac{dy}{dx} = \dfrac{d}{dx}\left( \dfrac{\ln(2)}{\ln(8)\ln(4)} + \dfrac{\ln(x)}{\ln(8)\ln(4)} \right) \\ \dfrac{dy}{dx} = \dfrac{d}{dx}\dfrac{\ln(2)}{\ln(8)\ln(4)} + \dfrac{d}{dx}\dfrac{\ln(x)}{\ln(8)\ln(4)} \\ \dfrac{dy}{dx} = \dfrac{d}{dx}\dfrac{\ln(2)}{\ln(8)\ln(4)} + \dfrac{1}{\ln(8)\ln(4)}\dfrac{d}{dx}\ln(x) \\ \dfrac{dy}{dx} = 0 + \dfrac{1}{\ln(8)\ln(4)} \cdot \dfrac{1}{x}\\ \dfrac{dy}{dx} = \dfrac{1}{\ln(8)\ln(4)x} \\$

9. Differentiate with respect to $x$: $y=\dfrac{\log_{2}(4x)}{\ln(8)}$

$\dfrac{d}{dx}y = \dfrac{d}{dx}\dfrac{\log_{2}(4x)}{\ln(8)} \\ \dfrac{dy}{dx} = \dfrac{d}{dx}\dfrac{\ln(4x)}{\ln(8)\ln(2)} \\ \dfrac{dy}{dx} = \dfrac{d}{dx} \dfrac{\ln(4) + \ln(x)}{\ln(8)\ln(2)} \\ \dfrac{dy}{dx} = \dfrac{d}{dx}\left( \dfrac{\ln(4)}{\ln(8)\ln(2)} + \dfrac{\ln(x)}{\ln(8)\ln(2)} \right) \\ \dfrac{dy}{dx} = \dfrac{d}{dx}\dfrac{\ln(4)}{\ln(8)\ln(2)} + \dfrac{d}{dx}\dfrac{\ln(x)}{\ln(8)\ln(2)} \\ \dfrac{dy}{dx} = \dfrac{d}{dx}\dfrac{\ln(4)}{\ln(8)\ln(2)} + \dfrac{1}{\ln(8)\ln(2)}\dfrac{d}{dx}\ln(x) \\ \dfrac{dy}{dx} = 0 + \dfrac{1}{\ln(8)\ln(2)}\cdot \dfrac{1}{x} \\ \dfrac{dy}{dx} = \dfrac{1}{\ln(8)\ln(2)x} \\$
10. Differentiate with respect to $x$: $y=5\log_{10}(x) - 10\log_{5}(x)$

$\dfrac{d}{dx}y = \dfrac{d}{dx}\left(5\log_{10}(x) - 10\log_{5}(x)\right)\\ \dfrac{dy}{dx} = 5\dfrac{d}{dx}\log_{10}(x) - 10\dfrac{d}{dx}\log_{5}(x)\\ \dfrac{dy}{dx} = 5\dfrac{d}{dx}\dfrac{\ln(x)}{\ln(10)} - 10\dfrac{d}{dx}\dfrac{\ln(x)}{\ln(5)}\\ \dfrac{dy}{dx} = \dfrac{5}{\ln(10)}\dfrac{d}{dx}\ln(x) - \dfrac{10}{\ln(5)}\dfrac{d}{dx}\ln(x)\\ \dfrac{dy}{dx} = \dfrac{5}{\ln(10)}\dfrac{1}{x} - \dfrac{10}{\ln(5)}\dfrac{1}{x}\\ \dfrac{dy}{dx} = \dfrac{5\ln(5)}{\ln(10)\ln(5)}\dfrac{1}{x} - \dfrac{10\ln(10)}{\ln(5)\ln(10)}\dfrac{1}{x}\\ \dfrac{dy}{dx} = \dfrac{5\ln(5)-10\ln(10)}{\ln(10)\ln(5)}\dfrac{1}{x}\\$