# Using definition of derivative w/ trig

1. Nov 4, 2007

### Cylord

1. I'm having trouble using the definition of derivative on algebraic functions involving trig functions. Here's the problem:

I need to find the first and second derivative of f(x) = $$x^{2}sin(x) + 2xsin(x)$$

2. I know how to use the definition of derivative, $$\stackrel{lim}{h->0}\frac{f(x+h) - f(x)}{h}$$, but I was never good at using it when trig functions were involved. I'm guessing the angle addition formula will be involved: sin(x+h) = sin(x)cos(h) + cos(x)sin(h) .

I also know that, if I can manage to get it in this form, $$\stackrel{lim}{h->0}\frac{cos(h) - 1}{h} = 0$$

3. I'm stuck at this point:
$$\stackrel{lim}{h->0}\frac{(x^2 + hx + h^2)(sin(x)cos(h) + cos(x)sin(h)) - x^2sin(x)}{h}$$

Any help would greatly be appreciated!

EDIT: I should also add that we haven't covered u-substition yet; we have to do this the long way without any shortcuts.

Last edited: Nov 4, 2007
2. Nov 4, 2007

### ZioX

Have you heard of l'hopitals rule? Joking...

Have you tried multiplying it all out and breaking it up into sums of fractions? It's messy, but it looks like it might work.

Remember your limit rules as well, if sin(h)/h -> 1 as h->0 then cos(x)sin(h)/h -> cos(x) as h->0.

Look at your x^2 terms: the coefficient is

$$\{\sin(x)\cos(h)+\cos(x)\sin(h)\}-\sin(x)=\sin(x)\{\cos(h)-1\}+\cos(x)\sin(h).$$

Last edited: Nov 4, 2007