Differentiation Rules

  1. Product Rule
  2. Chain Rule
  3. Reciprocal Rule
  4. Inverse-function Rule

Product Rule

To diffferentiate a product of functions, we apply the product rule:

`F(x)=u(x)v(x) rarr (dF(x))/(dx)=u(x) (dv(x))/(dx)+v(x)(du(x))/(dx)`

The method can be extended to products of more than two functions, for example, in the above example, were `v(x)=p(x)q(x)`, then we can apply the product recursively. We can state this as:

`(d(u(x)v(x)w(x)))/(dx)=u(x)v(x)(dw(x))/(dx)+u(x)w(x) (dv(x))/(dx)+w(x)v(x)(du(x))/(dx)`

Notice that each term in the sum contains the derivative of only one function, a different one in each.

Chain Rule

To differentiate a function of a function of x, we use the chain rule, which is integral to the method of u-substitution.

`(dF(G(x)))/(dx)= (dF(G(x)))/(dG(x)) (dG(x))/(dx)`

Reciprocal Rule

To differentiate a function which is a function of 1/x, for example,

`F(x)=1/(G(x)) rarr F(x)'=(-G(x)')/(G(x))^2`

Note, the Reciprocal rule is central to the Quotient-rule, which is not detailed here. For that see the Wikipedia article

Inverse-Function Rule

For a function of the form, `y=F(x)`, where `F(x)` has an inverse function such that `G(y)=x`, then

`(dy)/(dx)=(dF(x))/(dx)` and `(dx)/(dy)=(dG(y))/(dy)=H(y)`

In which case,

`(dy)/(dx)=(dy)/(dG(y))=1/(H(y))`

Example I:

Consider the function, `y=arcsin(x)` we have the inverse function `x=sin(y)`, so how do we go about finding `arcsin(x)'` ?

`(dy)/(dx)=(darcsin(x))/(dx)=??`

Using the inverse-function we have:

`(dx)/(dy)=(dsin(y))/(dy)=cos(y)`

On reciprocation we obtain,

`(dy)/(dx)=1/(cos(y))`

We can then solve this via one of two methods. Firstly, we can simply substitute in the definition `y=sin(x)`, to obtain

`(dy)/(dx)=1/(cos(arcsin(x)))`

Which can be expressed using the trigonometric relationship `cos(arcsin(z))=sin(arccos(z))=sqrt(1-z^2)`.

Alternatively, we can use Pythagoras theorum, `1=cos^2(z)+sin^2(z)`, solved for `cos(z)` ie.,

`cos(y)=sqrt(1-sin(y))`

Which we then simplify using the initial definition, `cos(y)=x` ie.,

`(dy)/(dx)=1/(cos(y))=1/sqrt(1-sin(y))=1/sqrt(1-x^2)`

Note, some authors prefer to employ the notation `F^(-1)(x)` to denote the inverse function. I would advise against this in most algebra and calculus (the exception being matrices) - always use the proper name of the inverse function! For example, if `sin^2(x)` is used to denote the function `sin(x)sin(x)`, then what is `sin^(-2)(x)` supposed to mean? Does it imply `1/sin(x) 1/sin(x)` ie., `csc^2(x)` OR `arcsin(x)arcsin(x)`? Avoid the ambiguity- use the proper name!

Example II:

Another good example of the use of the Inverse-function rule is in obtaining the derivative of `y=ln(x)`, where we use the inverse-function `x=exp(y)`.

`(dy)/(dx)=(dln(x))/(dx)=??` then `(dx)/(dy)=(dexp(y))/(dy)=exp(y)`

We substitute in the definition of `y`, and obtain,

`(dx)/(dy)=exp(ln(x))=x` and therefore `(dy)/(dx)=1/x`

 

 

See https://en.wikipedia.org/wiki/Chain_rule

https://en.wikipedia.org/wiki/Product_rule