# Mathematics for Chemistry/Differentiation

## Free web-based material from HEFCE[edit | edit source]

There is a DVD on differentiation at Math Tutor.

## The basic polynomial[edit | edit source]

The most basic kind of differentiation is:

There are two simple rules:

- The derivative of a function times a constant is just the same constant times the derivative.
- The derivative of a sum of functions is just the sum of the two derivatives.

To get higher derivatives such as the *second derivative* keep applying the same rules.

One of the big uses of differentiation is to find the stationary points of functions, the maxima and minima. If the function is smooth, (unlike a saw-tooth), these are easily located by solving equations where the first derivative is zero.

## The chain rule[edit | edit source]

This is best illustrated by example: find given

Let and .

Now and

So using the chain rule we have

## Differentiating a product[edit | edit source]

Notice when differentiating a product one generates two *terms*. (Terms are mathematical expression connected by a plus or minus.) An important point is that terms which represent physical quantities must have the same units and dimensions or must be pure dimensionless numbers. You cannot add 3 oranges to 2 pears to get 5 orangopears. Integration by parts also generates an extra term each time it is applied.

## Differentiating a quotient[edit | edit source]

You use this to differentiate .

### Problems[edit | edit source]

Differentiate with respect to

Notice we have .

Evaluate the inner brackets first.

Evaluate

a, b and c are constants. Differentiate with respect to .

### Answers[edit | edit source]

### Harder differentiation problems[edit | edit source]

Differentiate with respect to :

Differentiate with respect to

Differentiate with respect to

Evaluate

## Using differentiation to check turning points[edit | edit source]

is the tangent or gradient. At a minimum is zero. This is also true at a maximum or an inflection point. The *second gradient* gives us the nature of the point. If is positive the turning point is a minimum and if it is negative a maximum. Most of the time we are interested in minima except in *transition state theory*.

If the equation of is plotted, is is possible to see that at there is a third kind of point, an inflection point, where both and are zero.

Plot between -4 and +3, in units of 1. (It will speed things up if you factorise it first. Then you will see there are 3 places where so you only need calculate 5 points.) By factorising you can see that this equation has 3 roots. Find the 2 turning points. (Differentiate once and find the roots of the quadratic equation using . This gives the position of the 2 turning points either side of zero. As the equation is only in it has 3 roots and 2 maxima / minima at the most therefore we have solved everything. Differentiate your quadratic again to get . Notice that the turning point to the left of zero is a maximum *i.e.* and the other is a minimum *i.e.* .

What is the solution and the turning point of .

Solve , by factorisation.

(The 3 roots are -3,0 and +2.

Solutions are and , *i.e.* -1.7863 and 1.1196.

There are 3 coincident solutions at , , at 0 so this is an *inflection point*.

The roots are 0, 1 and -1.