A key part of any math students academic arsenal is the ability to find the derivative or a function. This is an invaluable skill when dealing with calculus and other higher level mathematics. Not only is being able to understand and preform differentiation, but it is a building block to integrals, another essential higher level math skill. Student’s are taught tricks in order to preform differentiation, such as the product rule, quotient rule, and chain rule, in order to obtain an answer quicker. This can leave many students feeling confused as to what exactly they are doing when they differentiate. To remedy this, I will discuss how to find the derivative of a function from first principles.
Consider the following general function y=f(x) (Note: y=f(x) could be ANY function.) Now consider this function at points x and x+h; the corresponding y values for these points are f(x) and f(x+h). Next, we consider the gradient of the line connecting these two points. From precious experience with coordinate geometry, we know that the gradient is equal to the difference in y divided by the difference in x.
Gradient = (difference in y)/(difference in x) = (f(x+h) – f(x)) / (x + h – x) = (f(x+h) – f(x)) / h
From a plot of f(x), we can see that as we make h smaller, we can more accurately measure the gradient of the curve at x. However, we can see from our general expression that we cannot simply let h be zero, so we instead take what is called the limit as h approaches zero. What this means is we simply look at the expression’s behavior as h becomes ever closer to zero. This can best be shown through example.
Example: y = x^2
Gradient = (f(x+h) – f(x)) / h = ((x + h)^2 – x^2 )/h = (x^2 +2xh + h^2 – x^2)/h = 2x + h
We observe h approaching zero, results in the gradient becoming 2x. This is the result we would expect which can be verified by performing one of the many rules for derivatives.
Example: y = x^3 when x=2
Gradient= (f(2+h) – f(2)) / h= ((4 + h)^3 – 4^3)/h= (4^3 +48h + 12h^2 + h^3 – 4^3)/h
= 48 + 12h + h^2
As h goes to zero, we can see that the gradient becomes 48, as required.
This article was written for you by Troy, one of the tutors with Test Prep Academy.