What is the Derivative of g(x) when x=0 in the Equation g(x) + x sin g(x) = x^2?

[chaslltt]

Homework Statement

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

Homework Equations

The Attempt at a Solution

At this point I have tried a few things but hit deadends. Any help would be appreciated.

 

[lanedance]

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

 

[Mark44]

Or just differentiate.

Show us what you have tried...

 

[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

 

[Mark44]

my first step took me to this:
g'(x) + xcos g'(x) + sin g(x) = 2x

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))?

 

[chaslltt]

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

 

[Mark44]

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

 

[chaslltt]

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

 

[Mark44]

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

 

[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)

 

[lanedance]

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

 

[blake knight]

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

 

[lanedance]

yeah i think we already outlined it

 

[blake knight]

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)

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

 

[blake knight]

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

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

 

[blake knight]

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

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.

 

[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.

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