User:Nutster/sandbox

A testing sandbox.

= Calculus Proofs =

Derivative Product Rule
$df(x)g(x)⁄dx = f '(x)g(x)+g'(x)f(x)$

Proof: $$\begin{align} \frac{df(x)g(x)}{dx} &= \lim_{h \to 0}{\frac{f(x+h)g(x+h)-f(x)g(x)}{h}} \\ &= \lim_{h \to 0}{\frac{f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)}{h}} \\ &= \lim_{h \to 0}{\frac{g(x+h) \left [ f(x+h)-f(x) \right ] + f(x) \left [ g(x+h)-g(x) \right ]}{h}} \\ &= \lim_{h \to 0}{\left ( \frac{g(x+h) \left [ f(x+h)-f(x) \right ]}{h} + \frac{f(x) \left [ g(x+h)-g(x) \right ]}{h} \right )} \\ &= \lim_{h \to 0}{\frac{g(x+h) \left [ f(x+h)-f(x) \right ]}{h}} + \lim_{h \to 0}{\frac{f(x) \left [ g(x+h)-g(x) \right ]}{h}} \\ &= \lim_{h \to 0}{g(x+h)} \lim_{h \to 0}{\frac{ f(x+h)-f(x) }{h}} + f(x) \lim_{h \to 0}{\frac{g(x+h)-g(x) }{h}} \\ &= g(x) \lim_{h \to 0}{\frac{ f(x+h)-f(x) }{h}} + f(x) \lim_{h \to 0}{\frac{g(x+h)-g(x) }{h}} \\ &= g(x) f'(x) + f(x) g'(x) \end{align}$$

Using Product rule
$$\begin{align} f(x) \div g(x) & = f(x) \left ( g(x) \right ) ^{-1} \\ \frac{df(x)\div g(x)}{dx} &= \frac{df(x) \left ( g(x) \right ) ^{-1}}{dx} \\ &= f'(x) \left ( g(x) \right ) ^{-1} + -1\left ( g(x) \right ) ^{-2} g'(x) f(x) \quad & \text{(Product rule, chain rule)} \\ &= \frac{f'(x)}{g(x)} - \frac{g'(x)f(x)}{g^2(x)} \\ &= \frac{f'(x) g(x)}{g^2(x)}- \frac{g'(x)f(x)}{g^2(x)} \quad & \left ( \text{Multiplied first fraction by } \frac{g(x)}{g(x)} \right )\\ &= \frac{f'(x) g(x) - g'(x)f(x) }{g^2(x)} \end{align}$$

Using the General Derivative
$$\begin{align} \frac{df(x)\div g(x)}{dx} &= \lim_{h \to 0}{\frac{f(x+h) \div g(x+h)-f(x) \div g(x)}{h}} \\ &= \lim_{h \to 0}{\frac{f(x+h) \div g(x+h)-f(x) \div g(x+h)+f(x) \div g(x+h)-f(x) \div g(x)}{h}} \\ &= \lim_{h \to 0}{\frac{\frac{1}{g(x+h)} \left [ f(x+h)-f(x) \right ] + f(x) \left [ \frac{1}{g(x+h)}-\frac{1}{g(x)} \right ]}{h}} \\ &= \lim_{h \to 0}{\left ( \frac{\frac{f(x+h)-f(x)}{g(x+h)}}{h} + \frac {f(x)}{h}\times\frac{g(x)-g(x+h)}{g(x+h)g(x)} \right ) }\\ &= \lim_{h \to 0}{\frac{f(x+h)-f(x)}{h \times g(x+h)}} + \lim_{h\to 0}{\left [ \frac{f(x)}{g(x+h)g(x)}\times \frac{g(x)-g(x+h)}{h} \right ] } \\ &= \lim_{h \to 0}{\frac{1}{g(x+h)}}\lim_{h \to 0}{\frac{f(x+h)-f(x)}{h}} + \lim_{h \to 0}{\frac{f(x)}{g(x+h)g(x)}}\lim_{h \to 0}{\frac{g(x)-g(x+h)}{h}}\\ &= \frac{1}{g(x)}f'(x) + \frac{f(x)}{g^2(x)} \left [-g'(x) \right ] \\ &= \frac{g(x)}{g^2(x)}f'(x) - \frac{f(x)}{g^2(x)} g'(x) \quad & \left ( \text{Multiplied first fraction by } \frac{g(x)}{g(x)} \right ) \\ &= \frac{f'(x) g(x) - g'(x)f(x) }{g^2(x)} \end{align}$$

d(x^x)/dx
$$\begin{align} x^x &= x^x \\ ln \left ( x^x \right ) &= ln \left ( x^x \right ) \\ &= x ln(x) \\ e^{ln(x^x)} & = e^{x ln(x)} \\ x^x &= e^{x ln(x)} \end{align}$$ $$\begin{align} \frac{dx^x}{dx} & = \frac{de^{x ln(x)}}{dx} \\ & = e^{x ln(x)} \times \frac{d x ln(x)}{dx} \\ & = e^{x ln(x)} \left [ 1 \times ln(x) + \frac{1}{x} x \right ] \\ & = x^x \left [ ln(x) + 1 \right ] \end{align}$$

d[x ln(x)]/dx
$$\begin{align} \frac {dxln(x)}{dx} & = x \times ln(x) + \frac{1}{x}x & \qquad ( \text{Product Rule} ) \\ & = x ln(x) + 1 \end{align}$$

Integral of ln(x) dx
$$\begin{align} & \text{Find } \int ln(x) dx \\ & \text{Let } u = ln(x), dv = 1 dx \\ & \therefore du = \frac{1}{x} dx, v = x \\ \end{align}$$

$$\begin{align} \int u\ dv & = uv - \int v\ du & \qquad ( \text{Integration by Parts } )\\ \int ln(x) dx & = x\ ln(x) - \int {x \frac{1}{x}} {dx} \\ & = x\ ln(x) - \int 1 dx \\ & = x\ ln(x) - x + C \\ \int ln(x) dx & = x \left [ ln(x)-1 \right ] + C \\ \end{align}$$