Posted: Feb 18, 2012 10:56 am
I will not have time to go through all the rest of the examples today; so I will just go through the second equation and solve this one in this installment.

The above is solved by seeing it as the sum of two terms; each term will be differentiated and their sum will be the first derivative of the above function. The same is true for the second derivative.

So we can say that:

and

then:

and the derivative will be:

Let's start with . In this case we have a=5, n=7, and n-1=7-1=6. So, the derivative will be:

This one was easy, if you follow the rules about differentiation I gave earlier. How about the second term? It is a negative power. Does that play any role in differentiating? The answer is no. Whether n is positive or negative it does not play any role at all in differentiating powers.

So for , we have a=1, n=-5, and n-1=-5-1=-6. Be careful with the signs here, as it is easy to make a mistake, and change the result. So, our result here is:

Putting the two derivatives together, we get:

And that's your result, for the first derivative.

The problem asks for the second derivative as well and here we have as terms:

So what is the second derivative of the first term? For we have a=35, n=6, n-1=6-1=5. Then this derivative will be:

What about the second term? Here we have a=-5, n=-6 and n-1=-6-1=7. Plugging in the numbers we have:

And the resulting second derivative will be:

Tomorrow, I will try and solve the other two examples.