Calculating Derivative of cos(xy)+ye^x Near (0,1) and Level Curve f(x,y)=f(0,1)

[Cpt Qwark]

Homework Statement

f(x,y)=cos(xy)+ye^{x} near (0,1), the level curve f(x,y)=f(0,1) can be described as y=g(x), calculate g'(0).

Homework Equations

N/A
Answer is -1.

The Attempt at a Solution

If you do f(0,1)=cos((0)(1))+1=2, do you have to use linear approximation or some other method?

 

[HallsofIvy]

Yes, at f(0, 1)= 2 so the level curve is cos(xy)+ e^xy= 2. Now, what they are calling "g(x)" is y(x) implicitly defined by that (that is, you could, theoretically, solve that equation for y.) Use implicit differentiation to find y'(x) from that.

 

[PWiz]

Why don't you first find an expression for $\frac{d(F(x,y))}{dx}$ ? You don't have to use any approximation methods here.

 

[Cpt Qwark]

So implicitly differentiate cos(xy)+e^{x}y=2?

I got something along the lines of \frac{dy}{dx}=\frac{ysin(xy)-ye^{x}}{-xsin(xy)+e^{x}}.

 

[PWiz]

So implicitly differentiate cos(xy)+e^{x}y=2?

I got something along the lines of \frac{dy}{dx}=\frac{ysin(xy)-ye^{x}}{-xsin(xy)+e^{x}}.

That is correct. Now substitute the $x$ and $y$ values and you'll get your answer. (Remember that for $f(0,1)=2$ , $g(x)=y$ . You've just found an expression for $g'(x)$.)