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 $g'(x)$.)