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Differentiation

Differentiation is part of calculus and can be used to find the rate of change. Useful applications of this in economics include being able to find the profit maximising point for firms by differentiating its costs and revenue (see here).

Rules of Differentiation
If given a function in x, you would differentiate to find dy/dx. That is y=ax differentiates to dy/dx=a (where a is a constant). The dy/dx simply means the differential of x. If you are given a different variable, for example you have y=at you would differentiate to find dy/dt=a (where a is a constant).

d[variable]/dx is a way of saying differentiate variable in terms of x. So for example d[5x+6]/dx means find the differential of (5x+6). If given f(x) then the differential is f'(x) (f dashed x) which is exactly the same as y differentiates to dy/dx just different notation.

Because we are finding the rate of change of variable there has to be a variable in the equation. For example we cannot just differentiate y=5000, because y is constant and there is no change, therefore we can only differentiate something that has a variable in it like y=500x. A constant differentiates to 0 (because it isnt changing).

Therefore if we differentiate y=5 we get dy/dx = 0.

We can differentiate many different powers to the x by using the rule that the differential of xn is nxn-1. That is we multiply x by the value of its power and then decrease the power by a value of 1. So if y=x2 then dy/dx=2x1 which is simply 2x.

Example
Differentiate y=2x, it may be helpful to include the power therefore y=2x1, hence dy/dx=2 (2*1x1-1 = 2x0 = 2). If we have to differentiate an equation that has more than one term, e.g. y=5x2 +6x, then we differentiate each term separately. Hence dy/dx = 10x + 6.

Differentiating Quotients
If you have to differentiate x where x is on the denominator then you would adjust the power to reflect this. Note 1/x is the same as x-1, and hence 1/x2 is the same as x-2. Differentiate y = 1/x - 5/x3; y=(x-1) - (5x-3) so dy/dx=(-x-2) - (-15x-4) which can also be written as dy/dx=(-1/x2)+(15/x4).

Factors
If you have to differentiate a factor then it is usually easier to expand the brackets and then differentiate. For example differentiate y = 5x(7x) is the same as y=35x2 and then differentiate to get dy/dx=70x. If the factor is a constant then this can be taken out and only the inside terms differentiated. For example differentiate y = 5(7x+6), this could be expanded or we could just do 5[d(7x+6)/dx] because the 5 is a constant. Therefore this equals 5 which equals 35, we would have got the same answer as if we had multiplied the 5 by the brackets.

Chain Rule
The chain rule can be used if you have brackets to the power of. For example (5x+6)3. We could either expand the brackets out (but this would take a long time) or use the chain rule. The chain rule is for y=(f(x))n then dy/dx=n(f(x))n-1 * (f'(x)). That is bring the power down in front of the f(x) and subtract a value of 1 from the power, and then multiply this by the differential of the brackets.
For example differentiate y= (7x+6)3; dy/dx = 3(7x+6)2 * (7) which can then be simplified and expanded.

Implicit Differentiation and Finding dy/dx from dx/dy
If you have the equation x=5y2 + 7y and you were to differentiate it you would get dx/dy, if you wanted to get it in the form of dy/dx then you can either differentiate implicitly or flip the equations. If we differentiate x=5y2 + 7y then we get dx/dy = 10y + 7, to get this in the form of dy/dx we can flip both sides. 10y+7 can also be written as (10y+7)/1. If we were to flip this we would get 1/(10y+7), if we were to flip dx/dy we would get dy/dx. Therefore dy/dx would equal 1/(10y+7).
Alternatively we could differentiate the equation implicitly. We have x = 5y2 + 7y and we want this in terms of dy/dx, to differentiate implicitly we would differentiate all terms like normal but since we want dy on top, any term that contains a y we would differentiate and also multiply by dy. So to differentiate x=5y2 + 7y implicitly we would get 1 = 10y.dy/dx + 7.dy/dx; note that I have used a dot as opposed to a multiplication symbol to avoid confusion with variables. To explain this we have differentiated x in terms of x (dx) to get 1, we have differentiated 5y2 to get 10y.dy/dx (remember we differentiate normally and then multiply any y terms (even if they are now to the power of zero; y0) by dy/dx) and we have differentiated 7y to get 7dy/dx. So we now have 1=10y.dy/dx + 7dy/dx; we would know factorise the dy/dx terms to get 1=dy/dx(10y+7), if we then divide both sides by (10y+7) we get dy/dx=1/(10y+7); note, this is the same as if we were to flip both sides. Hence both methods give the same answer, but sometimes implicit differentiation, although harder, has to be used.

Finding the 2nd Differential
You may be asked to find the second differential (expressed as d2y/dx2) this can be achieved by differentiating dy/dx in the normal fashion. For example find the d2y/dx2 of y=6x3+7x2+5x+4, dy/dx=18x2 + 14x + 5; d2y/dx2 = 36x+14.

Turning Point
If asked to find the co-ordinates of a turning point, we would differentiate an equation and set dy/dx equal to zero. We could then re-arrange the equation to find an x value, and then substitute this x value back into the initial equation to find the y value and hence we have found the co-ordinates of the turning point.
Example
For example find the co-ordinates of the turning point for the equation
y=7x2+2x
dy/dx = 14x+2
14x+2 = 0
14x=-2
x=-1/7
We then substitute this x value back into y=7x2+2x to find the y co-ordinate
y=7*(-1/7)2 + 2(-1/7)
y = -1/7
Therefore the co-ordinate of the turning point is (-1/7,-1/7)

Maximum or Minimum Point
We may also be asked to find out if this point is a maximum or minimum, to do this we find d2y/dx2 and substitute in the x value, if the result is a negative number then our result is a maximum point, if the result is positive then our result is a minimum point.
Continuing on from the example above, if we are asked to find out whether it is a maximum or a minimum point:
d2y/dx2 = 14
As this is a constant we have no need to substitute an x value in and because the result is positive we have found a minimum point.

Differentiation consists of much more than this, however this is the basics of differentiation and should be enough for simple mathematical economic applications. If you have any questions or queries about this please leave a message in the forum.

Try some of the questions below to ensure you understand differentiation.

Questions
1. Differentiate y = 2x + 6x2
2. Differentiate y = 5x6 + 7x2 + 5
3. Differentiate c = 1/t
4. Differentiate c = 7(x2 + 3)
5. Differentiate y = (t + 7)(t + 2)
6. Differentiate f(x) = (6x2 + 5)7
7. Differentiate x = 6y4 + 5y2 + 2y3 + 3 in terms of dy/dx (you can either use the flip method or differentiate implicitly)
8. Differentiate p = 12q2 + 7 in terms of dq/dp  