MATH 120 - Day 9

2024-10-02 13:13:11 -0400 EDT

Simple derivative Proofs

Proof: $f(x) = c \Rightarrow f’(x) = 0$

Let $ f = c$

$ \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h} = \frac{c - c}{h} = \frac{0}{h} = 0$

The limit exists and is finite for every $ x \in \mathbb{R}$, so $f$ is differentiable on $\mathbb{R}$ and $f’(x) = 0$

Proof: $f(x) = x^n \Rightarrow f’(x) = nx^{n-1}$

Let $ f = x^n$

$ \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h} = \frac{(x + h)^n - x^n}{h} = \frac{x^n + nx^{n-1}h + 2nx^{n - 2}h^2 … h^n - x^n}{h} = \frac{nx^{n-1}h + 2nx^{n - 2}h^2 … h^n}{h}$
$ = nx^{n-1} + 2nx^{n - 2}h … h^{n-1}$, let h = 0
$ = nx^{n-1}$

The limit exists and is finite for every $x \in \mathbb{R}$, so $f$ is differentiable on $\mathbb{R}$ and $f’(x) = nx^{n-1}$.

Proof: $f(x) = e^x \Rightarrow f’(x) = e^x$

Let $ f = e^x$

$ \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h} =\frac{e^{(x + h)} - e^x}{h} = \frac{e^x (e^h - 1)}{h} = e^x \cdot \frac{e^h - 1}{h}$
$ = \lim\limits_{h \to 0} e^x \cdot \lim\limits_{h \to 0} \frac{e^h - 1}{h} = \lim\limits_{h \to 0} e^x \cdot 1 = e^x$

$\lim\limits_{h \to 0} \frac{e^h - 1}{h} = 1$, (just trust me on this. its really hard to explain)

The limit exists and is finite for every blah blah blah $f’(x) = e^x$.

Proof: $f(x) = \sin x \Rightarrow f’(x) = \cos x$

Let $ f = \sin x$

$ \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h} = \frac{ \sin{(x + h)} - \sin{(x)}}{h} = \frac{ \cos{x} \sin{h} + \sin{x} \cos{h} - \sin{(x)}}{h}$
$ = \frac{ \cos{x} \sin{h}}{h} + \frac{ \sin{x} \cos{h} - \sin{(x)}}{h} = \cos{x} \frac{ \sin{h}}{h} - \sin{x} \frac{(1 - \cos{h})}{h}$
$ = \cos{x} (1) - \sin{x} (0) = \cos{x}$

Proof: $f(x) = \cos x \Rightarrow f’(x) = - \sin x$

This works almost exactly the same as the derivative of sine; sub in a cos, the last part of the equation switches and the result if $ - \sin{X}$

Proof: $(f + g)’ = f’ + g’ $

Let $ F = (f + g)$

$ \lim\limits_{h \to 0} \frac{F(x + h) - F(x)}{h} = \frac{f(x + h) + g(x + h) - f(x) - g(x)}{h} = \frac{f(x + h) - f(x)}{h} + \frac{g(x + h) - g(x)}{h} = f’ + g'$

Note: The proof for subtraction works exactly the same

Proof: $(f \cdot g)’ = f’ \cdot g + f \cdot g'$

Let $ F = (f \cdot g) $

$ \lim\limits_{h \to 0} \frac{F(x + h) - F(x)}{h} = \frac{f(x + h)g(x + h) - f(x)g(x)}{h}$
$ = \frac{f(x + h)g(x + h) - f(x)g(x) + f(x)g(x + h) - f(x)g(x + h)}{h}$ $ = \frac{g(x + h)(f(x + h) - f(x)) + f(x)(g(x + h) - g(x))}{h} = g(x + h) \frac{f(x + h) - f(x)}{h} + f(x) \frac{g(x + h) - g(x)}{h}$
$ = g(x + h) f’ + f(x) g’ = g(x) f’ + f(x) g'$

Proof: $(cf)’ = c \cdot f'$

Let $ F = c \cdot f $

$ \lim\limits_{h \to 0} \frac{F(x + h) - F(x)}{h} = \frac{c \cdot f(x + h) - c \cdot f(x)}{h} = c \cdot ( \frac{f(x + h) - f(x)}{h}) = c \cdot f'$

$ F’ = c \cdot f'$

Proof: $(\frac{f}{g})’ = \frac{f’ \cdot g + f \cdot g’}{g^2}$

Let $ F =(\frac{f}{g}) $

$ \lim\limits_{h \to 0} \frac{F(x + h) - F(x)}{h}$

$ = \frac{\frac{f(x + h)}{g(x + h)} - \frac{f(x)}{g(x)}}{h}$

$ = \frac{\frac{f(x + h)g(x) - f(x)g(x + h)}{g(x + h)g(x)}}{h}$

$ = \frac{1}{g(x + h)g(x)} \cdot ( \frac{f(x + h)g(x) - f(x)g(x + h)}{h})$

$ = \frac{1}{g(x + h)g(x)} \cdot ( \frac{f(x + h)g(x) - f(x)g(x + h) + f(x + h)g(x + h) - f(x + h)g(x + h)}{h})$

$ = \frac{1}{g(x + h)g(x)} \cdot (g(x + h) \cdot \frac{f(x + h) - f(x)}{h} - f(x + h) \cdot \frac{g(x + h) - g(x)}{h})$

$ = \frac{1}{g(x + h)g(x)} \cdot (g(x + h) \cdot f’ - f(x + h) \cdot g'$

$ = \frac{1}{g(x)g(x)} \cdot (g(x) \cdot f’ - f(x) \cdot g'$

$ = \frac{1}{g^2} \cdot (f’ \cdot g - f \cdot g'$

$ = \frac{f’ \cdot g - f \cdot g’}{g^2}$

Proof: $\frac{d}{dx}(g(f(x)) = \frac{dg}{dx} = \frac{dg}{df} \cdot \frac{df}{dx} = f’(g(x)) \cdot g’(x)$

Let $ F = g(f(x))$

$ \lim\limits_{h \to 0} \frac{F(x + h) - F(x)}{h}$
$ \frac{g(f(x + h)) - g(f(x))}{h}$
$ \frac{g(f(x + h)) - g(f(x))}{h} \cdot \frac{f(x + h) - f(x)}{f(x + h) - f(x)}$
$ \frac{g(f(x + h)) - g(f(x))}{f(x + h) - f(x)} \cdot \frac{f(x + h) - f(x)}{h}$

Let t = f(x + h), u = f(x)

$ \lim\limits_{t \to u} \frac{g(t) - g(u)}{t - u} \cdot f’(x)$
$ g’(u) \cdot f’(x) = g’(f(x)) \cdot f’(x)$