Calculus II - closed function question

[axcelenator]

there is a function: F ( x, y, z) = 2ln (xz) + sin ( xyz) − y^2 = 0.
the func is defined by the closed function z=f(x,y) and provides : f(1,0)=1
we define: g(t)=f(t,1-t^6) . where t is very close to 1.
I have to find g'(1)

Homework Equations

I tried to to do like that: find F'x and F'z and did z'x =-(F'x/ F'z) and got -1. but from here I don't know what to do.
The answer is g'(1)=2.
Thanks for your help!

 

[lanedance]

sorry, can't really follow this - can you explain the question in a little more detail and more clearly? for example what is g in relation to F and f and x in relation to t?

 

[axcelenator]

sorry about the wrong copying: g(t)=f(t,1-t^6)

 

[HallsofIvy]

there is a function: F ( x, y, z) = 2ln (xz) + sin ( xyz) − y^2 = 0.
the func is defined by the closed function z=f(x,y) and provides : f(1,0)=1

Are you saying there exist f(x,y) such that z= f(x,y), in the neighborhood of (1, 0, 1), is the same as 2ln(xz)+ sin(xyz)- y^2= 0?

we define: g(t)=f(t,1-t^6) . where t is very close to 1.
I have to find g'(1)

If so, then F(x, y, g)= 2ln(xg)+ sin(xyg)- y^2= 0. And with x= t, y= 1- t^6, that is 2ln(tg)+ sin(t(1- t^6)g)- (1- t^6)^2.

Differentiating with respect to t,
\frac{2}{tg}(1+ tg')+ cos(t(1- t^6)g)((1- t^6)g- 6t^6g+ t(1-t^6)g')- 12(1- t^6)t^5= 0
Set t= 1, so that x= 1, y= 0, and F(1, 0, g(1))= 2ln(g(1))= 0. What is g(1)? Put x= 1, y= 0, and that value of g(1) into the equation and solve for g'(1).

Homework Equations

I tried to to do like that: find F'x and F'z and did z'x =-(F'x/ F'z) and got -1. but from here I don't know what to do.
The answer is g'(1)=2.
Thanks for your help!