# List of Notes - Category: Differentiation Rules

## Constant Multiple Rule

The Constant Multiple Rule. If c is a constant anf f is a differentiable function then (cf(x))'=c (f(x))' .

Proof. By definition (cf(x))'=lim_(h->0)(cf(x+h)-cf(x))/h=c lim_(h->0)(f(x+h)-f(x))/h=cf'(x).

## Sum and Difference Rules

The Sum Rule. If f and g are both differentiable then (f(x)+g(x))'=(f(x))'+(g(x))'.

Proof. By definition (f(x)+g(x))'=lim_(h->0)((f(x+h)+g(x+h))-(f(x)+g(x)))/h=

=lim_(h->0)((f(x+h)-f(x))/h+(g(x+h)-g(x))/h)=

## Product Rule

Product Rule. If f and g are both differentiable then (f(x)g(x))'=f(x)g'(x)+f'(x)g(x).

Proof.

By definition

(f(x)g(x))=lim_(h->0)(f(x+h)g(x+h)-f(x)g(x))/h=

=lim_(h->0)(f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x))/h=

## Quotient Rule

Quotient Rule. If f and g are differentiable then ((f(x))/(g(x)))'=(f'(x)g(x)-f(x)g'(x))/(g(x))^2.

Proof.

Let h(x)=(f(x))/(g(x)) then f(x)=h(x)g(x).

Differentiating both sides gives: f'(x)=(h(x)g(x))' or using product rule f'(x)=h(x)g'(x)+h'(x)g(x).

## Chain Rule

Now let's see how to differentiate composite functions.

Suppose that we are given function h(x)=f(g(x)). Remembering that g'(x) is rate of change of g(x) with respect to x and f'(g(x)) is rate of change of f with respect to g(x) then it is reasonable to suggest that rate of change of f with respect to x is product of f'(g(x)) and g'(x).

## Implicit Differentiation

The techniques we learned allows us to differentiate functions like y=root(3)(x^2+sin(x)) or y=tan(x)/(e^(sin(x))) or, in general, y=f(x). But what to do if we can't express y in terms of x, or expression will be very complex?

## Derivative of Inverse Function

Now, let's use implicit differentiation to find derivatives of inverse functions.

Fact. Suppose that function y=f(x) has unique inverse and has finite derivative f'(x)!=0 then derivative of inverse function y=f^(-1)(x) is (f^(-1)(x))'=1/(f'(f^(-1)(x))).

## Logarithmic Differentiation

The calculation of derivatives of functions involving products, powers or quotients can be simplified with the logarithmic differentiation (because of the properties of logarithms).

Let's first see how to differentiate functions that already have product and/or quotient under logarithm.