# Calculating derivative

1. May 23, 2015

### Cpt Qwark

1. The problem statement, all variables and given/known data
$$f(x,y)=cos(xy)+ye^{x}$$ near $$(0,1)$$, the level curve $$f(x,y)=f(0,1)$$ can be described as $$y=g(x)$$, calculate $$g'(0)$$.

2. Relevant equations
N/A
3. 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?

2. May 23, 2015

### HallsofIvy

Staff Emeritus
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.

3. May 23, 2015

### 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.

4. May 23, 2015

### Cpt Qwark

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

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

5. May 23, 2015

### PWiz

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 $g'(x)$.)