First Principals Question

[jp1010]

I'm trying to get the derivative of f(x)=(x^3)cos(1/x^2) from first principals
I know that the derivative is 3x^2cos(1/x^2)+2sin(1/x^2)
I can see where the 3x^2cos(1/x^2) comes from and where the other parts cancel but i cant seem to get the 2sin(1/x^2) part.
Any suggestions/hints would be much appreciated, thanks.

 

[Steven G]

First principals? I would assume that this would mean to find the derivative using the definition of derivative. However I doubt that is what you meant.
So where are you stuck? What have you tried? If you show us your work then we can see where you are making your mistake.

I would use the product rule for this problem. Please show us what you get using the product rule and then we will help you from there.

 

[jp1010]

First principals? I would assume that this would mean to find the derivative using the definition of derivative. However I doubt that is what you meant.
So where are you stuck? What have you tried? If you show us your work then we can see where you are making your mistake.

I would use the product rule for this problem. Please show us what you get using the product rule and then we will help you from there.

from the product rule i get 3x^2cos(1/x^2)+2sin(1/x^2) which i know to be correct. I want to show that this is the derivative from first principals.
i.e from f'(x)=lim h->0 (f(x+h)-f(x))/h
for the function f(x)=x^3cos(1/x^2)
f'(x)=lim h->0 (((x+h)^3)cos(1/(x+h)^2)-x^3cos(1/x^2))/h
from here i am very stuck. from similar questions the double angle formula is often used which would result in the 2sin(1/x^2) that i need. But i cannot find a way to use it that would result in the correct derivative.

 

[Dr.Peterson]

We have theorems (such as the product rule) for a reason! They allow us to break a complicated problem (such as finding a derivative) into manageable pieces, so that we never have to do what you are trying to do.

If you try to apply the definition of the derivative to your complicated function directly, it will be extremely difficult; in the process, you will find yourself redoing what is done in proving the various theorems that are used in differentiating the "easy" way -- not only the product rule, but the chain rule, the derivatives of powers and cosines, and more.

It just isn't worth the effort, unless you were assigned this as an exercise with a very high point value. If you just want to do this for the experience, start with something smaller, such as x cos(x) or cos(1/x), and work up to the big one. But the main thing you would learn from the exercise is never to do it again. (I learned that lesson long ago.)

 

[HallsofIvy]

Well, using "first principles", i.e. using the definition of the derivative, is tedious but I will give it a shot.
The derivative of y= f(x), at \(\displaystyle x= x_0\) is defined as \(\displaystyle \lim_{h\to 0}\frac{f(x_0+h)- f(x_0)}{h}.

Here, \(\displaystyle f(x_0)= x_0^3cos(1/x_0^2)\) and \(\displaystyle f(x_0+ h)= (x_0+ h)^3cos(1/(x_0+ h)^2)= (x_0^3+ 3hx_0^2+ 3h^2x_0+ h^3)cos(1/(x_0^2+ 2x_0h+ h^2)\). It's that \(\displaystyle cos(1/(x_0^2+ 2x_0h+ h^2)\) that is the hard part! I think you will need to convert that \(\displaystyle 1/(x_0^2+ 2x_0h+ h^2\) to an infinite series then use the Taylor series expansion of cosine to convert the whole thing to a power series. Fortunately you can ignore terms that have powers of h greater than 1.

Who in the world assigned such a horrid problem? As Dr. Peterson said, this should give you a great respect for derivative formulas!\)