Now lets see how we can differentiate some simple algebraic expressions.
Case 1: Let's begin with the simple expression y = x^2. Now remember the fundamental notion about calculus is the idea of growing. Mathematicians call it varying. Now as y and x^2 are equal to each other, it is clear that if x grows, x^2 will also grow. And if x^2 grows, then y will also grow. What we have got to find out its the proportion between the growing of y and the growing of x. In other words our task od to find out the ratio between dy and dx, in brief, to find the value of dy/dx.
Let x, grow a little bit larger and become x + dx; similarly, y will grow a bit bigger and will become y + dy. THen clearly, it will still be true the enlarged y will be equal to the square of the enlarged x. Writing this down we have:
y + dy = (x + dx)^2.
Doing the squaring we get: y + dy = x^2 + 2x*dx + (dx)^2
What does (dx)^2 mean? Remember that dx meant a little bit of x. Then (dx)^2 will mean a little bit of a little bit of x; that is, as explained in the previous post it is a small quantity of the second order of smallness. It may be discarded as quite inconsiderable in comparison with the other terms. Leaving it out, we then have
y + dy = x^2 + 2x*dx.
Now y = x^2; so we will subtract it from the the equation above and we have left
dy = 2x*dx.
Multiplying by dx on both sides: dy/dx = 2x.
Now this is what we set out to find. he ratio of the growing of y to the growing of x is found to be 2x.
If we put numbers tot the above example, it might make more sense. Suppose x = 100, therefore y = 10,000. hen let x grow until it becomes 101, (that is, dx = 1). Then the enlarged y will be 101 * 101 = 10,201. But if we agree that we may ignore small quantities of the second order, 1 may be rejected as compared with 10,000; so we may round off the enlarged y to 10,200. The bit added on is dy, which is therefore 200.
dy/dx = 200/1.
According to the algebra-working above, we find dy/dx = 2x. And so it is; for x = 100, 2x = 200.
