Monday, June 20, 2011

6.2 – Derivatives of a Function Defined Implicitly or Parametrically

You probably have learnt how to differentiate and integrate functions implicitly and parametrically, but only up to the first order. Here, we will be learning how to continue on to the 2nd order. It is actually very easy and straight-forward, so there is nothing too important in this section.


I think I don’t need to tell you how to do it. differentiating a function implicitly for 2nd order is just the same as 1st order. I’ll show you an example:

Find the 2nd order derivative of the function x2 + y2 = 2.

Note the use of the product rule in this question. Just do more exercises, then you will get used to these kind of questions.


Probably there’s something new in this section. Again, I’ll show you an example:

Consider the parametric equations x = t + 1 and y = t3.
Differentiating each other with respect to t gives

To find d2y/dx2,

But we cannot differentiate 3t2 with respect to x. Therefore, using chain rule,

To summarize it up, finding the 2nd order derivative for parametric equations x and y is by the equation:

This section acts as a revision to what you are learning in Maths T. Please get very familiar with differentiation and integration before continuing to the next section.

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