# Derivative using chain rule

1. Apr 8, 2008

### goldfronts1

I am trying to find the first and second derivative using the chain rule of the following:

u sin(x^2)

This is what I have but I don't think it is correct. Can someone pls let me know?

first derivative: u * 2x cos(x^2) + sin(x^2) u'

second derivative:
u * 2( x * -2sin(x^2) + cos(x^2)) + 2xcos(x^2)* u' + sin(x^2)*u" + u'* 2xcos(x^2)

Any help thanks

2. Apr 8, 2008

### jamesrc

That's pretty close: the first term (for your second derivative) is not correct but I think if you take a second look you will figure it out (you just missed one detail). Otherwise, you can combine like terms to clean it up a bit, but it's still a long expression.

3. Apr 8, 2008

### goldfronts1

I don't get it :(

4. Apr 8, 2008

### sutupidmath

i am assuming u is an other function right?

Think of it this way

$$[(u2x)cos(x^2)]'$$ in other words think first of (u2x) as a single function, so you can differentiate this using the product rule for (u2x) and
cos(x^2) right? after that apply again the product rule wherever you end up with (u2x)' and you'll be fine.

5. Apr 8, 2008

### goldfronts1

This is what I get

2ucos(x^2)-4x^2sin(x^2)+4xcos(x^2)u'+sin(x^2)u"

Is this correct?
Thanks

6. Apr 8, 2008

### sutupidmath

You have this question on another post as well, and it looked fine to me as well there.
Well i take this back, you are short of an x in there.
you should have -2xsin(x^2) somewhere at the beggining, on your original result, at fpost #1.

Last edited: Apr 8, 2008
7. Apr 8, 2008

### sutupidmath

This looks fine.