Differentiation Formulae

Here I summarise the recursive rules for deriving (differentiating) compound expressions.

There are many different notations for differentiation. I will mainly use Lagrange’s here; although it is less precise than some others, it will make what follows concise.

So:

f' = \dfrac{df}{dx} = \dfrac{d}{dx}f(x) = f_x = D_x\,f = D\,f

I will also write

(f \circ g)(x) = f (g (x)),
i.e. composition of functions with a circle

(f \cdot g)(x) = (fg)(x) = f(x) \times g(x),
i.e. multiplicative product of functions with a dot or juxtaposition;

these are both products but in different senses.

Note that mathematical notation can be a bit ambiguous or inconsistent. For example:

f^{-1} is the inverse function (compositionally), not the reciprocal (which is written 1/f); but

f^2(x) = (f(x))^2 is the (multiplicative) square of the result of the function.

Fundamental Rules

This is a nearly minimal set of basic rules. Such rules are usually established from limit arguments on the definition of derivation.

[const] c' = 0, (i.e. for constant with respect to x);

[id] x' = 1;

[+] (f+g)' = f'+g';

[×] (fg)' = f'g + fg';

[o] (f \circ g) = (f' \circ g) \cdot g';

[^] (f^g)' = f^g \left(\dfrac{f'g}{f} + g' \ln(f) \right);

\sin' = \cos;
\cos' = - \sin;
\sinh' = \cosh;
\cosh' = \sinh;

(e^x)' = e^x;

(\ln|x|)' = 1/x.

More Rules

From these a little algebra may be used to obtain:

[~] (-h)' = -\,h'
(from [const,+]: 0 = 0' = (h + (-h))' = h' + (-h)');

[1/] \left(\dfrac{1}{h}\right)' = - \dfrac{h'}{h^2}
(from [const,×]: 0 = 1' = (h(1/h))' = h'(1/h) + h(1/h)' = h'/h + h(1/h)');

[{\Box}^{-1}] (h^{-1})' = \dfrac{1}{h' \circ h^{-1}};
(from [const,o]: 1 = x' = (h \circ h^{-1})' = (h' \circ h^{-1}) \cdot (h^{-1})').

These in turn lead to:

[-] (f-g)' = f'-g';

[÷] (f/g)' = \dfrac{f'g - fg'}{g^2}.

Yet More Rules

These all lead to, for example:

\tan' = (\sin/\cos)' = (\sin' \cos - \sin \cos') / \cos^2
= (\cos^2 + \sin^2) / \cos^2 = 1/\cos^2 = 1 + \tan^2;

(x^a)' = x^a \left(\dfrac{x'a}{x} + a' \ln(x) \right) = x^a \left(\dfrac{a}{x} + 0 \ln(x) \right) = ax^{a-1};

(a^x)' = a^x \left(\dfrac{a'x}{a} + x' \ln(a) \right) = a^x \left(\dfrac{0}{a} + \ln(a) \right) = a^x \ln a;

(\sinh^{-1}(x))' =\dfrac{1}{\sinh' \circ \sinh^{-1}(x)} =\dfrac{1}{\cosh (\sinh^{-1}(x))} = \dfrac{1}{\sqrt{x^2+1}}.

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