Differentiation

Linear approximation

Tangent line is used to approximate a function $f(x)$ near a point. The slope $m$ of the tangent line at a point $a$ is the derivative of the function $f(x)$ at that point:

$$ m_a = f^\prime (a) $$

Let $x$ be any point near the initial point $a$ that we are interested in. Then $\Delta x = x - a$. Since $x$ is near $a$, then $\Delta x \to 0$. The difference in the value of the function $\Delta f = f(x) - f(a)$ is approximately qual to:

$$ \Delta f \approx m_a \cdot \Delta x $$

$$ = \frac{df}{dx} \bigg|_ {x=a} \cdot \Delta x \quad \text{ for } \Delta x \text{ near } 0 $$

Finally, to find an approximate value of $f(x)$ we add the value $f(a)$ to the approximate change $\Delta f$:

$$ f(x) \approx \Delta f + f(a) = f^{\prime}(a)(x-a) + f(a) $$

The second derivative of a function can tell us if the function is concave up or concave down, so we can estimate if the approximation is higher than the actual value or lower than it.

Product rule

Imagine a ractangle whose sides are concurently expanding, so that the area is increasing. The rate of change of the area of such a rectangle represents a derivative of the product of two functions. Let $h(x) = f(x) \cdot g(x)$, then $\Delta h$ is equal to the new value of the function minus the original value of the function:

$$ \Delta h = (f + \Delta f)(g + \Delta g) - f \cdot g $$

$$ = \cancel{f \cdot g} + f \cdot \Delta g + \Delta f \cdot g + \Delta f \cdot \Delta g - \cancel{f \cdot g} $$

$$ = f \cdot \Delta g + \Delta f \cdot g + \Delta f \cdot \Delta g $$

In order to find the derivative of $h(x)$, we need to find the limit of $\frac{\Delta h}{\Delta x}$ as $\Delta x \to 0$:

$$ \frac{d}{dx} h(x) = \frac{d}{dx} \bigg( f(x) \cdot g(x) \bigg) = \lim_{\Delta x \to 0} \frac{\Delta h}{\Delta x} $$

$$ = \lim_{\Delta x \to 0} \bigg( f \cdot \frac{\Delta g}{\Delta x} + \frac{\Delta f}{\Delta x} \cdot g + \frac{\Delta f}{\Delta x} \frac{\Delta g}{\Delta x} \Delta x \bigg) $$

$$ = f \cdot g^\prime + f^\prime \cdot g + f^\prime \cdot g^\prime \cdot 0 = f^\prime \cdot g + g^\prime \cdot f $$

at all points where derivatives $f^\prime (x)$ and $g^\prime (x)$ are defined.

Quotient rule

Let $h(x) = \frac{f(x)}{g(x)}$, then $\Delta h$ is equal to the new value of the function minus the original value of the function:

$$ \Delta h = \frac{f + \Delta f}{g + \Delta g} - \frac{f}{g} $$

$$ = \frac{\cancel{f \cdot g} + \Delta f \cdot g - \cancel{f \cdot g} - f \cdot \Delta g}{g(g + \Delta g)}$$

$$ = \frac{\Delta f \cdot g - f \cdot \Delta g}{g(g + \Delta g)}$$

In order to find the derivative of $h(x)$, we need to find the limit of $\frac{\Delta h}{\Delta x}$ as $\Delta x \to 0$:

$$ \frac{d}{dx} h (x) = \frac{d}{dx} \frac{f(x)}{g(x)} = \lim_{\Delta x \to 0} \frac{\Delta h}{\Delta x} $$

$$ = \lim_{\Delta x \to 0} \frac{\frac{\Delta f \cdot g - f \cdot \Delta g}{g(g + \Delta g)}}{\Delta x} = \lim_{\Delta x \to 0} \frac{\frac{\Delta f}{\Delta x} \cdot g - \frac{\Delta g}{\Delta x} \cdot f}{g(g + \Delta g)} $$

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

at all points where derivatives $f^\prime (x)$ and $g^\prime (x)$ are defined and $g(x) \ne 0$

Why $\Delta g \to 0$? Let's apply the precise definition of the limit. Suppose that $\Delta x = x-a$ and $\Delta f = f(x)-L$. Here $x$ is a variable and it changes together with $f(x)$ while $a$ and $L$ are constants. Therefore, the precise definition states that $\lim_{x \to a} f(x) = L$, which means that for a given $\epsilon \gt 0$ we can find $\delta \gt 0$ such that when $0 \lt |\Delta x| \lt \delta$, then $|\Delta f| \lt \epsilon$ (provided that $f(x)$ is one to one in the interval). No matter how small $\epsilon \gt 0$ we are given, we can still find $\delta \gt 0$ so that when $\Delta x$ is within $\delta$ of $a$, it implies that $\Delta f$ is within $\epsilon$ of $L$, in other words, as $\Delta x \to 0$, then $\Delta f \to 0$.

Derivatives of all trigonometric functions

$$ \frac{d}{dx}\sin x = \cos(x) $$

$$ \frac{d}{dx}\cos x = -\sin(x) $$

$$ \frac{d}{dx}\tan x = \frac{1}{\cos^2 x} = \sec^2 x $$

$$ \frac{d}{dx}\cot x = -\frac{1}{\sin^2 x} = -\csc^2 x $$

$$ \frac{d}{dx}\sec x = \frac{\sin x}{\cos^2 x} = \sec x \tan x $$

$$ \frac{d}{dx}\csc x = -\frac{\cos x}{\sin^2 x} = -\csc x \cot x $$

Chain rule

If $h(x) = f(g(x))$, then

$$ h'(x) = f'\left(g(x)\right) g'(x) $$

at all points where the derivatives $f'(g(x))$ and $g'(x)$ are defined.

Alternatively, if $y = f(u)$, and $u = g(x)$, then

$$ \left. \frac{dy}{dx} \right|{x = a} = \left. \frac{dy}{du} \right|{u=g(a)} \left. \frac{du}{dx} \right|_ {x=a} $$

The derivative of an inverse function

If $g$ is a (full or partial) inverse of a function $f$, then

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

at all $x$ where $f'(g(x))$ exists and is non-zero.

The derivative of the exponential function

$\displaystyle \frac{d}{dx} a^x = \lim_{\Delta x \to 0} \frac{a^{x + \Delta x} - a^x}{\Delta x}$ $\displaystyle = \lim_{\Delta x \to 0} a^x \frac{a^{\Delta x} - 1}{\Delta x}$ $\displaystyle = a^x \lim_{\Delta x \to 0} \frac{a^{\Delta x} - 1}{\Delta x}$

Let $\displaystyle M(a) = \lim_{\Delta x \to 0} \frac{a^{\Delta x} - 1}{\Delta x}$, then

$$ \frac{d}{dx} a^x = M(a) a^x $$

We define $e$ to be a unique real number, so that $M(e) = 1$. Then

$$ \frac{d}{dx} e^x = M(e) e^x = e^x $$

Because $\Delta x$ is moving, $a^x$ is constant which we can factor out.

$$ \left. \frac{d}{dx} a^x \right|_ {x = 0} = M(a) a^0 = M(a) $$

$M(a)$ is the slope of the graph of $a^x$ at $x = 0$

Derivative of the natural logarithm function

$$ \displaystyle \frac{d}{dx} \ln x = \frac{1}{x} $$

Change to base e

$$ \displaystyle \frac{d}{dx} a^x = ? $$

$$ \displaystyle a^x = \left( e^{\ln a} \right)^x = e^{x \ln a}$$

$$ \frac{d}{dx} a^x = \frac{d}{dx} e^{x \ln a} = (\ln a) e^{x \ln a}$$

$$ \frac{d}{dx} a^x = (\ln a) a^x $$

From the previous formula, we can find that $\displaystyle M(a) = \ln a$

Logarithmic differentiation

$$ \frac{d}{dx} u = ?$$

Sometimes it is easier to differentiate the logarithm of a given function than the function itself.

$$ \frac{d}{dx} \ln u = \left( \frac{d \ln u}{du} \right) \left( \frac{du}{dx} \right) = \frac{1}{u} \frac{du}{dx} $$

$$ \left( \ln u^{\prime} \right) = \frac{u^{\prime}}{u} $$

$$ \frac{d}{dx} a^x = ? $$

Let $u = a^x$, then take the logarithm from both sides:

$$ \ln u = x \ln a $$

$$ (\ln u)^{\prime} = \ln a $$

$$ \frac{u^{\prime}}{u} = \ln a $$

$$ u^{\prime} = u \ln a$$

$$ \frac{d}{dx} a^x = (\ln a) a^x $$

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Mean Value Theorem

If $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there is a point $c$ in $(a,b)$, such that the average rate of change over $[a,b]$ is equal to the instantaneous rate of change at $c$.

$$\frac{f(b)-f(a)}{b-a}=f^\prime(c)$$

In mathematics, a hypothesis is a part of a propositional statement. Hypothesis may or may not be true. Generally, we want to explore what happens if we assume that the hypothesis is true.