Differentiation basics
Differentiation Basics
Differentiation is a tool of mathematics that is primarily used for calculating rates of change.
It is basically the difference between two points on a line with respect to some independent variable. On a straight line, the differentiation is the gradient of that straight line (see figure 1). Since the gradient of a straight line is constant over the entire line, then the differentiation is also the same. However for a curve, the gradient at a point is found by drawing a tangent to that point and then finding the gradient of that line (figure 2). As a result, the gradient changes as we move around the curve.
Figure 1. Straight line graph
In real life, differentiation is applied in virtually all strata. When you are discussing the change of price of a commodity, you are basically differentiating. Say the price of biscuits was £3 last week and is now £3.20 7 days later, there has been a 20p change over a week. Therefore, we can say that the rate of change of the price of biscuit is
£0.20/7days
That is, change in price divided by the number of change.
When we differentiate, we are simply comparing the change in a particular variable with another variable.
Figure 2 (figure courtesy s-cool.co.uk)
In this article, we will be introducing the concept of differentiation.
Differentiation by rule
Usually in differentiation, it is common to use y and x as our two variables. The notation is many instances is
dy/dx
where the ‘d’ stands for ‘change in’. Therefore the above can be read as
change in y/ change in x
In some instances in mathematics, we can write y as a function of x
y = f(x)
In this case, differentiation of y is written as
y’ = f’(x)
or
y’ = df/dx
The ‘prime’ indicates that it is the first derivative of the variable f. In other words, it means we are differentiating f once.
Note that any letter can be used in place of y, f and x. However, these are probably the most common letters used.
The rule for differentiation
Next step is to find out, how do we differentiate?
Given a function
y = axn
where y and x are the variables and a and n are constants, to differentiate y above, we simply apply the formula below
dy/dx = (a * n) xn-1
Simple isn’t it?
The question could also be written as
d(axn)/dx
for which the solution remains the same ie.
d(axn)/dx = (a * n) xn-1
Let’s try it in an example. Suppose we want to differentiate
y = 2×3
This will give us
dy/dx = (2*3) x3-1
dy/dx = 6×2
One more example
Find
As before
d(5x^(-1) )/dx=(-1*5) x^(-1-1)
The appropriate form for differentiation
One more example. This example is set to explain a very important point to note before differentiating.
Find
How do we solve this as the equation does not look like one we’ve dealt with before. The key here is to try and transform into a form we are used to i.e. axn.
Luckily for us, we can use the law of indices to sort this out.
Rearranging the above
4/√x=4/x^(1/2)
That’s the first step. It’s not looking like axn yet but it’s the first step in a two-step process. Let’s go to step 2
Step 2
4/x^(1/2) =〖4x〗^(-1/2)
Now this looks more like it where a = 4 and n = -1/2
Let’s now differentiate
= (-1/2*4) x^(-1/2-1)= -2x^(-3/2)
And that’s it! Problem solved. If you like, you can rearrange the above to give
d(4/√x)/dx= -2/√(x^3 )
One more thing to note before we move on to applications. What is we want to differentiate a constant? So for example, differentiate the equation
y = 5
We can re-write the above as
y = 5×0
afterall, x0 is 1
This differentiating as before
y = (0 * 5)x0-1
y = 0x-1
y = 0
In general, the differentiation of any constant is 0. Keep this in mind.
In the next section, we will look one application of differentiation.
Tangents to curves
As mentioned earlier, we can find the gradient at a point of a curve. Doing this requires that we find a tangent at that point and then find the gradient of the tangent. What we can also do with differentiation is to find the equation of the tangent itself.
Mathematically, given a curve whose equation is
y = 2×2 + 3x + 4
To find the equation of the tangent at point where x = 3 for example, we follow the steps below
Step 1
Find the y coordinate when x is 3.
y = 2(32) + 3(3) + 4
y = 2(9) + 9 + 4
y = 18 + 9 + 4
y = 31
Therefore, the coordinate of the point is (3, 31)
Step 2
Differentiate the equation of the curve
y’ = (2 * 2)x2-1 + (1 *3)x1-1 + 0
y’ = 4x + 3
Step 3
We now substitute the value of x = 3 into the equation y’
y’ = 4(3) + 3
y’ = 12 + 3
y’ = 15
This is the gradient of the line at point when x = 3 and y = 31
Now that we know the gradient, we can then go ahead to find the equation of this tangent
Step 4
Remember when we found the change in price of biscuits with respect to the number of days earlier in this article? When, we use the same principle next. We know the gradient (say m) is
m = Change in y/change in x
m = (y – y1) / (x – x1)
We know the initial values of x (x1) and y (y1) (i.e. 3 and 31) and the value of the gradient. So inserting these into the equation above, we have
15=(y_ -31)/(x_ -3)
Cross multiplying
15(x_ -3)=y_ -31
15x_ -45=y_ -31
15x-y= -31+45
15x – y = 14
This is the equation of the tangent to the curve at point x = 3.
If we rearrange this equation,
y = 15x – 14
and compare this with the general equation of a straight line
y = mx + c
where m is the gradient, we will find that m in this case is 15 which is the same answer derived above.
It looks like a long process but if the steps above are followed, it should lead you straight to the solution.
Let us use our knowledge of gradients, tangents and equation of a straight line to solve a question.
Example
Suppose we want to find the point(s) at which the curve with equation is parallel to the line with equation
Step 1
Find a function for the gradient for any point on the curve. This is done simply by differentiating the equation of the curve.
y’ = 10x – 2
Step 2
Find the gradient of the straight line.
Since we already know that the general equation of a straight line is y = mx + c, we first rearrange the equation of the line to look like our general equation so that we can easily find the the gradient of the line.
4x – 3y = 7
Can be re-written as
y = (7-4x) /-3
or
y = -7/3 + 4x/3
Straight forward, we can see that the gradient is the coefficient of x which is 4/3.
Now we know that for the curve, we can always calculate the gradient using the derivative
y’ = 10x – 2
We also know that at a particular point on the curve, the gradient of the tangent to that point is -4/3. To find that point, we simply equate the derivative to the gradient
y’ = 10x – 2 = 4/3
10x = 4/3 + 2
10x = 10/3
x = 1/3
When x is 1/5, then we can solve for y by substituting into the equation of the curve
y = 5(1/3)2 – 2(1/3) + 3
y = 5(1/9) – 2/3 + 3
y = 5/9 +7/3
y = 26/9
Therefore, the point at which the curve with equation is parallel to the line with equation is
(1/3, 26/9)
Increasing and decreasing functions
Finally we will be looking at increasing and decreasing functions.
Let us say we have a curve whose equation is
y = 3x – x2
The derivative of this curve is
y’ = 3 – 2x
This gives us the equation for the gradient at any point on the graph. So for example, at point where x = 1 and y = 2, the gradient is
y’ = 3 – 2(1) = 2
This gradient is a positive gradient. When the tangent has a positive gradient, we call this an increasing function.
As you may have guessed, when the tangent has a negative gradient, we call it a negative function.
For example, at point when x = 3 and y = 0, the gradient is
y’ = 3 – 2(3) = -3
So at point x = 3, we say it’s a decreasing function.
Figure 3 Graph of y = 3x – x2
So what about when the gradient is zero as in the case when x = 3/2?
y’ = 3 – 2(3/2) = 0
In this case, the tangent to this point is parallel to the x axis and the function is neither increasing nor decreasing and hence is known as a stationary point. Because the function is increasing before reaching the stationary point and decreasing just after the stationary point, it is known as a maximum point. If the function was decreasing before the stationary point and then increasing after the stationary point, then that stationary point is known a minimum point. The graph in figure 3 above has a maximum stationary point at x = 1.5 and he graph in figure 2 has a minimum stationary point at x = 0.
Asides the maximum and minimum turning point, there is one more type of stationary point called the point of inflexion. This is the case where the function is increasing (decreasing) before the stationary point and continues to increase (decrease) after the stationary point as shown in figure 4.
Figure 4. Point of inflexion shown in graph of y = x3
Let’s try an example to round this off.
Example
For g(x) = x3 – 2x, determine
whether g is increasing or decreasing at x = 0
whether g is increasing or decreasing at x = 1
the value(s) of x for which g is stationary.
First we differentiate g(x)
g^’ (x)=〖3x〗^2-2
Then we substitute x = 0 into this derivative
g^’ (x)=〖3(0〗^2)-2= -2
Since g’(x) is negative, then the function is decreasing at x = 0
When x = 1
g^’ (x)=〖3(1〗^2)-2= 1
In this case, g’(x) is positive therefore the function is increasing at x = 1
At what point is g stationary? Remember, g is stationary when g’(x) = 0
Or
g^’ (x)=3x^2-2=0
3x^2=2
x^2=2/3
x= ±√(2/3)
This implies that there are two stationary points on the curve. One occurs at
x= √(2/3)
And the other at
x= -√(2/3)
This is shown in the figure below
Figure 5. Graph of g(x) = x3 – 2x
