# Find the derivative of the function(chain rule)

1. Aug 4, 2015

### youngstudent16

1. The problem statement, all variables and given/known data
Various values of the functions f(x) and g(x) and their derivatives are given in the table below. Find the derivative of f(x+g(x)) at x=0.

at x=0 f(x)=5 f'(x)=2 g(x)=1 g'(x)=3
at x=1 f(x)=7 f'(x)=3 g(x)=-2 g'(x)=-5
2. Relevant equations
Chain rule

3. The attempt at a solution

So f'(x)(x+g(x))[(1+g'(x))(f(x)]
2(0+1)(1+3)(5)=40

Is this correct or did I overlook something chain rule always has a way of making me feel I missed something

2. Aug 4, 2015

### Lok

Could you detail the chain rule used? Just the base formula.

3. Aug 4, 2015

### youngstudent16

Derivative of the outside multiplied by the derivative of the inside
(f*g)'=(f'*g)+(g'*f)

Where f=f(x)
g=(x+g(x))

I'm getting 40 but not sure if I attempted this correctly

4. Aug 4, 2015

### Lok

I asked as this did not seem the right chain rule use. If i understood the problem correctly.
One issue is that f(x+g(x)) is not equal to f*(x+g(x)).
f(u) is a function and in your case u=x+g(x).
What chain rule do you need to apply in this case?

5. Aug 4, 2015

### Qwertywerty

This has been written wrong .

d( f(x+g(x)) )/dx = f'(x+g(x))*(1+g'(x)) .

6. Aug 4, 2015

### Lok

Ahhhh. Much better.
And now, just solve :P

7. Aug 4, 2015

### youngstudent16

f'(x+g(x))
How do I solve this with the given information? Would it just be the derivative f a constant which makes the whole thing 0?

8. Aug 4, 2015

### Lok

Well you don't know anything about the function f, but solving for x=0 does just mean f'(0+g(0)).

9. Aug 4, 2015

### Qwertywerty

10. Aug 4, 2015

### youngstudent16

I would plug in the value for x and g(0) getting derivative of 1 which is 0. Making the final answer 0.

I think thats right I really confused myself on the chain rule

11. Aug 4, 2015

### Lok

but f'(1) is not zero by the conditions above.

12. Aug 4, 2015

### Lok

Sorry about the mixed responses to both of you. I have been switching through screens all morning.

13. Aug 4, 2015

### Qwertywerty

You don't find derivative by putting the value before differentiating .

14. Aug 4, 2015

### youngstudent16

oh wow that x=1 did come in handy I got it finally 12 was correct

thanks for the patience really helped with my confusion

15. Aug 4, 2015

### Qwertywerty

My suggestion to you is that you revise differentiation once again .

Last edited: Aug 4, 2015
16. Aug 4, 2015

### Lok

Anytime there is any time :P.