# Differentiation problem

1. Oct 13, 2009

### chaslltt

1. The problem statement, all variables and given/known data

If g(x) + x sin g(x) = x^2 find g'(0)

2. Relevant equations

3. The attempt at a solution
At this point I have tried a few things but hit deadends. Any help would be appreciated.

2. Oct 13, 2009

### lanedance

try using implict differentiation, then sub in for x = 0

3. Oct 13, 2009

### Staff: Mentor

Or just differentiate.

Show us what you have tried...

4. Oct 13, 2009

### chaslltt

my first step took me to this:
g'(x) + xcos g'(x) + sin g(x) = 2x
then i set it equal to g'(x)
g'(x)= 2x- sin g(x)/xcos(1)
I would then plug in 0 but that makes the denominator 0

5. Oct 13, 2009

### Staff: Mentor

The first, third, and fourth terms are correct, but the second is wrong. To differentiate x sin(g(x)) you need the product rule and the chain rule.
d/dx(x sin(g(x))) = x*d/dx(sin(g(x)) + 1* sin(g(x))

d/dx(sin(g(x)) is NOT cos(g'(x)). That's not how the chain rule works.

What should you get with d/dx(f(g(x))?

6. Oct 13, 2009

### chaslltt

d/dx f(g(x)) would be f'(x)g(x)*g'(x)

7. Oct 13, 2009

### Staff: Mentor

Yes. Now what is d/dx(sin(g(x))?

8. Oct 13, 2009

### chaslltt

cos(g(x))*g'(x)

9. Oct 13, 2009

### Staff: Mentor

Yes. Now put this back your differentiation problem for the part that was incorrect.

10. Oct 13, 2009

### chaslltt

alright i ended up getting
g'(x)= 2x- sin g(x)/(1+cos g(x)
so then i plug in the 0 but what is g(0)

11. Oct 13, 2009

### lanedance

have a look at your original equation, when x = 0

12. Oct 13, 2009

### blake knight

not quite my friends...there is a solution to this one. dig more.

13. Oct 13, 2009

### lanedance

yeah i think we already outlined it

14. Oct 14, 2009

### blake knight

First of all theres something missing in this equation: the x in front of cos g(x).

15. Oct 14, 2009

### blake knight

You should obtain an equation in the form of: g'(x)=[2x-sin g(x)]/[1+xcos g(x)].

16. Oct 14, 2009

### blake knight

Now, understanding what the problem really asks for, it is asking for g'(0), meaning what is the value of g'(x) when x=0.

17. Oct 14, 2009

### blake knight

You can substitute 0 now for x, you will end up getting: g'(0) = -sin g(x)<----this solves the problem.