Question

Differentiate with respect to x cos²x³

Answer 1

Implicit Differentiation

This section covers Implicit Differentiation.

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  =  3y2
dy

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    3y2
In 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 ax with respect to x.

You might be tempted to write xax-1  as the answer. This is wrong. That would be the answer if we were differentiating with respect to a not x.

Put y = ax .

Then, taking logarithms of both sides, we get:

ln y = ln (ax)
so ln y = x lna

So, differentiating implicitly, we get: (1/y) (dy/dx) = lna
and so dy/dx = y lna = ax lna