Intergration

If y³ = x, how would you differentiate this with respect to x? There are three ways:

Method 1

Rewrite it as y = x^(1/3) and differentiate as normal (in harder cases, this is not possible!)

Method 2

Find dx/dy:
dx  =  3y²dy

And now use the fact:	dy	=	1
	dx		dx/dy

So we get:
dy  =  1
dx     3y²
Method 3 Differentiate term by term and use the chain rule:
y³   =   x

d	(y3)	=	d	(x)
dx			dx

The right hand side of this equation is 1, since the derivative of x is 1. However, to work out the left hand side we must use the chain rule.
The left hand side becomes:
d (y³) ×  dy
dy         dx
(although it is not strictly correct to do so, at this level you can think of dy/dx as a fraction in the chain rule. In the line above, imagine that you can cancel the "dy" s, leaving d/dx and y³, which is what we had in the previous line).
Therefore, 3y² × dy  = 1
                         dx
So   dy  =  1
       dx    3y²In this example, method (2) is probably the easiest. However, there are cases when the only possible method is (3).

Example

Differentiate x² + y² = 3x, with respect to x.

d	(x2)	+	d	(y2)	=	d	(3x)
dx			dx			dx

2x + d (y²)×dy  =  3
       dy        dx
2x + 2y dy  = 3
           dx
dy  =  3 - 2x
dx        2y

Example

Differentiate a^x with respect to x.
You might be tempted to write xa^x-1  as the answer. This is wrong. That would be the answer if we were differentiating with respect to a not x.
Put y = a^x .
Then, taking logarithms of both sides, we get:
ln y = ln (a^x)
so ln y = x lna
So, differentiating implicitly, we get: (1/y) (dy/dx) = lna
and so dy/dx = y lna = a^x lna
(copy)  ::D
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