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I have to use implicit differentiation to find $\frac{dy}{dx}$ given:

$$x^2 \cos(y) + \sin(2y) = xy$$

I don't even know where to begin, I missed the class where we went over implicit differentiation, and because of that, I am completely stuck.

Thank you everyone.

Edit: I don't know how to make the equation look all nice and whatnot, so sorry about that

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Have you looked at… ? It gives a description and some examples. – Ross Millikan May 16 '11 at 14:01
To make the equations pretty, you enclose $\LaTeX$ in dollar signs. If you right click on any example and select "Show Source" you can see how it was done. – Ross Millikan May 16 '11 at 14:03
If you are you familiar with partial derivatives you can differentiate with respect to $x$ the function $F(x,y)=x^2\cos(y)+\sin(2y)-xy\equiv 0$. – Américo Tavares May 16 '11 at 16:09

Differentiating w.r.t $x$ we have $$2x \cos{y} - x^2\sin{y}\frac{\text dy}{\text dx} + 2\cos{2y}\frac{\text dy}{\text dx} = y + x \frac{\text dy}{\text dx}$$

From the above you have \begin{align*} 2x \cos{y} - y &= x \cdot\frac{dy}{dx} + \sin{y}\cdot\frac{dy}{dx} -2\cos{2y}\cdot\frac{dy}{dx} \\ &= \frac{dy}{dx} \cdot \Bigl[ x + \sin{y} - 2\cos{2y}\Bigr] \end{align*}

Take all the $\frac{\text dy}{\text dx}$ to one side and then simplify. And look at this Wikipedia Link. You have some worked out examples.

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An equation $F(x,y)=0$ defines a curve $\gamma$ in the $(x,y)$-plane. In most cases (as here) it is impossible to solve the equation for $y$ explicitly. But one can say the following: Any point $(x_0,y_0)\in\gamma$ where $F_y(x_0,y_0)\ne 0$ is the center of a rectangular window $W$, such that $\gamma \cap W$ is the graph of a ("local") function $y=\phi (x)$. One may ask for the derivative $\phi'$, in particular for its value at $x_0$. It is shown in the calculus course you missed that this value can be calculated ${\it without\ formally\ solving\ the\ equation}$ $F(x,y)=0$. The result is $$\phi'(x_0)=-{F_x(x_0,y_0)\over F_y(x_0,y_0)}\ .$$ Doing the calculation you will end up with an expression containing both variables $x$ and $y$. If $(x_0,y_0)\in\gamma$ is given in advance you may just plug these coordinates in. If not, your derivative $\phi'$ appears not as a function of $x$ alone but as a function of $x$ and $y\ (=\phi(x))$. In addition the resulting expression is not uniquely determined insofar as it may be simplified using the equation $F(x,y)=0$.

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Expanding my comment above. Assume you have a differentiable implicit function $F(x,y)=0$. Let $y=f(x)$ denote the function such that $F(x,f(x))\equiv 0$. This function $f(x)$ does not need to be explicitly known, as pointed out in Christian Blatter's answer.

If you differentiate both sides of $F(x,y)=0$ and apply the chain rule, you get the following total derivative with respect to $x$:

$$\frac{\mathrm{d}F}{\mathrm{d}x}=\frac{\partial F}{\partial x}+\frac{% \partial F}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}x}\equiv 0.\qquad (1)$$

Solving $(1)$ for $\frac{\mathrm{d}y}{\mathrm{d}x}$, gives you the following formula

$$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{\partial F}{\partial x}/\frac{% \partial F}{\partial y}.\qquad (2)$$

For $F(x,y)=x^{2}\cos (y)+\sin (2y)-xy$, since the partial derivatives are

$$\frac{\partial F }{\partial x}=2x\cos y-y\quad\text{ and }\qquad \frac{\partial F }{\partial y} =-x^{2}\sin y+2\cos 2y-x,$$

the derivative of $y=f(x)$ is

$$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{2x\cos y-y}{-x^{2}\sin y+2\cos 2y-x}.$$

As a final note I say that the way I presented the computation is the same as it is explained in Advanced Calculus book by Angus Taylor. I have seen this computation done here many times without considering partial derivatives explicitly, but it is essential the same as when evaluating those partial derivatives.

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The idea is that $y$ is a function of $x$. Therefore, to differentiate terms like $f(y)$, one uses the chain rule:

$$ \frac{d}{dx}(f(y)) = f'(y) \frac{dy}{dx}. $$

One also must use the product rule to differentiate the product of two or more terms (things like $x^2 y$ or $x^5 \sin(y)$ or $x \sin(xy^2)$, etc.).

Along with the comments and links above, this should be more than enough to get you on your way.

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