Find the derivative of the function(chain rule)

[youngstudent16]

Homework Statement

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. Homework Equations
Chain rule

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 [/B]

 

[Lok]

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

 

[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

 

[Lok]

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

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?

 

[Qwertywerty]

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

This has been written wrong .

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

 

[Lok]

This has been written wrong .

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

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

 

[youngstudent16]

This has been written wrong .

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

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?

 

[Lok]

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?

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

 

[Qwertywerty]

Would it just be the derivative f a constant which makes the whole thing 0?

I don't follow you .

 

[youngstudent16]

I don't follow you .

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 that's right I really confused myself on the chain rule

 

[Lok]

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 that's right I really confused myself on the chain rule

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

 

[Lok]

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

 

[Qwertywerty]

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

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

 

[youngstudent16]

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

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

 

[Qwertywerty]

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

thanks for the patience really helped with the my confusion

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

 

[Lok]

Anytime there is any time :P.