Differentiation - "First Principles"

Question

Differentiate the function f(x) = 5x^2-4x+7 using first principle.

Answer 1

f(x) = 5x^2 -4x + 7

f''(x) =2(5)x^(2-1) -1(4)x^(1-1) +0(7)= 10x -4 by short cut, multiply each term by the exponent, and reduce the exponent by 1

which is the same derivative as

differentiation with first principles

f'(x) = limit as h approaches 0 of [f(x+h) - f(x)]/h

f(x+h) = 5(x+h)^2 -4(x+h) +7 = 5x^2 +10xh +5h^2 -4x-4h + 7

f(x)= 5x^2 -4x +7

f(x+h)-f(x) = 5x^2 +10xh+5h^2 -4x -4h +7 - 5x^2 +4x -7 = 10xh +5h^2 -4h

[f(x+h)-f(x)] /h = (10xh+5h^2-4h)/h = 10xh/h +5h^2/h -4h/h = 10x +5h -4

limit as h approaches 0 of 10x+5h-4 = 10x+5(0)-4 = 10x+0-4

= 10x-4

Answer 2

The idea behind a derivate is that for a lot of purposes we want to find the slope at a point. Let's step back and look at how we'd find the slope between two points- let's say we wanted to find the slope between (0,1) and (2,5).

Slope is defined as rise divided by run. In this case, the rise would be the difference in the y values-so 5-1=4. The run would be the difference in the x values, which would be 2-0=2. We'd then divide our rise (4) by our run (2) and get a slope of 4/2=2.

Suppose instead of being given the y values we were instead given y(x)=x²+1 and asked to find the slope between x=0 and x=2. In this case, we'd do the same thing as above, except we'd have to work out what y(0) and y(2) are.

Now we're ready to talk about differentiation. Derivates are usually thought of as the slope of a function, not between two points, but between only one point. Suppose we wanted to find the derivative of y(x) at x=2. If we tried the same method of rise divided by run, we'd get 5-5 divided by 2-2, or 0 divided by 0-something that makes mathematicians very angry, so we don't want to do that.

Instead, let's take a look at what happens when we take the slope between x=2 and numbers getting progressively closer and closer to 2. For example, the slope of y(x) between x=2 and x=1 is 5-2 divided by 2-1, or 3 divided by 1, which is 3. Using the same method, we could get the slope between x=2 and x=1.9 to be 3.9. Between x=2 and x=1.99, we'd get a lope of 3.99, between x=2 and x=1.999 we'd get a slope of 3.999, and so on. The closer we take the lower number to 2, the closer the slope gets to 4.

Now we're finally ready to learn about the first principle of calculus.

In practice, this can be a very tedious thing to do every time we need to calculate a derivative. What we instead do is define the variable h to be the difference between the point we want to find the derivative of and the the other point that we're using to find the slope, and then look at what happens when h becomes very small.

The slope between x=2 and x=2+h would then be ((2+h)²+1)-(2²+1) divided by (2+h)-2.

(2+h)²+1 can be expanded to 2*2+2*h+h*2+h*h+1=5+4h+h², so the rise would be 5+4h+h²-5=4h+h². The run would then be 2+h-2=h.

Dividing the rise by the run, we would get 4+h. If h is really, really small, 4+h becomes really, really close to 4-which would be our derivative.