Differentiation

Differentiation

Let us look at hunting for this dydx for some simple equations.

Consider y=x2

If x grows, x2 growsdecreases

But x2=

So if x grows y growsdecreases

What we have got to find out is the proportion between the growing of y and the growing of x. ie dydx when dx is really really small.

Let x, then, grow a little bit bigger and become x+dx; similarly, y will grow a bit bigger and will become y+dy.

That is we have:

y+dy=x+dx2

y+dy=x2+2x⋅dx2x^2+dx2

What does dx2 mean?

Remember that dx meant a little bit–of .

Then dx2 will mean a little bit of a little bit of x;

So dx2 is a small quantity of the secondfirstthird order of smallness.

Consider the equation below. Click on the part of the equation which can be discarded because it is very small.

y+dy=x2+2x·dx+dx2

After dropping the very very small dx2 we have

y+dy=x2+2x·dx

But y=x2. So we have

dy=2x·dx

Dividing across by dx we get:

dydx=

Now this is what we set out to find. The ratio of the growing of y to the growing of x is, in the case before us, found to be 2x.

Definition

This ratio dydx is the result of differentiating y with respect to x.

What about the equation u=7x2+3

If we were told to differentiate this with respect to x, we should have to find dudydx

=d7x2+3dx

Let time be the independent variable, for example in y=b+12at2. Then, if we were told to differentiate it, that means we must find its differential coefficient with respect to

So that then our business would be to try to find dydtdx

That is, to find db+12at2dt

Given y=x2. Suppose x=100 and y=10,000. Let x grow till it becomes 101 (that is, let dx=).

Then the enlarged y will be .

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. y has grown from 10,000 to

The bit added on is dy, which is therefore

dydx=/

dydx=200=2·100=2·x

dydx=2x when y=x2

Try differentiating y=x3 in the same way.

We let y grow to y+dy, while x grows to x+.

Then we have y+dy=x+dx3.

Doing the cubing we obtain y+dy=x3+3x2·dx+3xdx2+dx3.

Now we know that we may neglect small quantities of the second and third orders. Click the equation parts which can be discarded below.

y+dy=x3+3x2·dx+3xdx2+dx3

So, regarding them as negligible, we have left: y+dy=x3+3x2·dx.

But y=x3; and, subtracting this, we have: dy=3x2*

and dy/=3x2.

dydx=3x2 when y=x3

Try differentiating y=x^4. Starting as before by letting both y and x grow a bit, we have: y+dy=(x+)^4.

Working out the raising to the fourth power, we get

y+dy=x4+4x3dx+6x2dx2+4xdx3+dx4

Then striking out the terms containing all the higher powers of dx, as being negligible by comparison. Click the equation parts which can be discarded below.

y+dy=x4+4x3dx+6x2dx2+4xdx3+dx4

We have y+dy=x4+4x3dx.

Subtracting the original y=x4, we have left

dy=4x3*

and dy/=4x3.

dy/dx=4x3 when y=x4

Now all these cases are quite easy. Let us collect the results to see if we can infer any general rule.

Put them in two columns, the values of y in one and the corresponding values found for dy/dx in the other:

| y | dy/dx |

| x2 | |

| x3 | 3x23x3 |

| x4 | 4x34x4 |

Can you see a pattern forming?

For y=xn, it seems that dydx = nxyx^n-1n+2.

The general rule appears to be: When y=xn, dydx=nxn−1.

Case of a negative power.

Let y=x−2. Then proceed as before:

y+dy=x+dx−2=x^*1+dxx−2.

Expanding this by the binomial theorem, we get:

=x−21−2dxx+22+11·2dxx2−etc.

=x−2−2x−3·dx+3x−4dx2−4x−5dx3+etc.

So, neglecting the small quantities of higher orders of smallness, we have:

y+dy=x−2−2x−3·dx.

Subtracting the original y=x−2, we find

dy=−2x−3*

and dy/dx=-2x^-3-4.

This is still in accordance with the rule inferred above: For y=xn, dydx=nxn−1, where n = .

Case of a fractional power.

Let y=x12. Then, as before,

y+dy=x+dx12=x121+dxx12

=x+12dxx−18dx2x·x+terms with higher powers of dx.

Subtracting the original y=x12, and neglecting higher powers we have left:

dy=12dxx=12x−12·dx,

and dy/dx=12x−12.

This agrees with the general rule: For y=xn, dydx=nxn−1, where n = .

Summary:To differentiate xn, multiply by the power and reduce the power by one, so giving us nxn−1 as the result.