Respuesta :

Differentiating implicity gives 

g'(x) + xcos(g(x))(g'(x)) + sin(g(x)) = 2x 

g'(x)[1 + xcos(g(x))] = 2x - sin(g(x)) 

g'(x) = [2x - sin(g(x))] / [1 + xcos(g(x))] 

g'(0) = -sin(g(0))


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facundo3141592

We want to find the g'(0) by knowing that:

g(x) + x*sin(g(x)) = x^2

We will get:

g'(0) = 0

First, we can see that:

g(0) is given by:

g(0) + 0*sin(g(0)) = 0^2

g(0) = 0.

Now let's derive this, we can rewrite:

g(x) = x^2 - x*sin(g(x))

Now we just differentiate to get:

g'(x) = 2*x - sin(g(x)) - x*cos(g(x))*g'(x)

Now we need to move all the terms with g'(x) to the left side:

g'(x) + x*cos(g(x))*g'(x) = 2x - sin(g(x))

g'(x)*(1 +  x*cos(g(x)) =  2x - sin(g(x))

g'(x) = ( 2x - sin(g(x)) )/(1 +  x*cos(g(x))

Now we just need to evaluate this in x = 0, so we get:

g'(0) = (2*0 - sin(g(0)))/(1 + 0*cos(g(0)) = -sin(g(0))

And remember that g(0) = 0 then:

g'(0) = -sin(0) = 0

g'(0) = 0

If you want to learn more, you can read:

https://brainly.com/question/19573890

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