I searched and couldn't find anything specific to my question, so I'll ask it here.

I'm asked to find the indicated derivative:

$${\operatorname{d}y\over\operatorname{d}x} \sin(xy^2)-x^2 = x+5$$

What's throwing me off is the direction to solve in terms of x and y. I've gone through my notes and can't find anything regarding finding a derivative for different terms.

Any assistance in setting this up is greatly appreciated.

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    $\begingroup$ I'm a bit confused... is the derivative multiplied by the sine, or are you being asked to find $y^\prime$ if $\sin(xy^2)-x^2 = x+5$? In the latter case, one merely uses the chain rule; e.g. for the sinusoidal term the derivative is $\cos(xy^2)(2xyy^\prime+y^2)$ $\endgroup$ – J. M. is a poor mathematician Apr 18 '11 at 21:54
  • $\begingroup$ Look up the Implicit Function Theorem. $\endgroup$ – Yuval Filmus Apr 18 '11 at 22:00
  • $\begingroup$ Or, just look up implicit differentiation, assuming you meant what J.M. said. Implicit Function Theorem is probably a bit much for someone probably in Calc 1. $\endgroup$ – Graphth Apr 18 '11 at 22:08

Taken literally, from what you wrote: $$\frac{dy}{dx}\sin(xy^2) - x^2 = x+5$$ you find $\frac{dy}{dx}$ "in terms of $x$ and $y$" simply by moving $x^2$ to the other side and then dividing through by $\sin(xy^2)$: $$\begin{align*} \frac{dy}{dx}\sin(xy^2) - x^2 &= x+5\\ \frac{dy}{dx}\sin(xy^2) &= x^2 + x + 5\\ \frac{dy}{dx} &= \frac{x^2+x+5}{\sin(xy^2)}. \end{align*}$$

However, I suspect this is not what your problem is. It is unfortunate that you refer to instructions but don't quote them; when you are confused by the statement of a problem, it is best to quote it and then say what confuses you. I fear your confusion has caused you to misreport what the problem actually says.

I suspect that your problem says for you to your find $\frac{dy}{dx}$, in terms of $x$ and $y$, if $$\sin(xy^2) - x^2 = x+5.$$

This is called implicit differentiation. This equation defines $y$ implicitly as a function of $x$: given any value of $x$, you plug it in, and you find the values of $y$ that make the equation true. Since $y$ is a function of $x$ (though only implicitly), you can ask what the derivative of $y$ with respect to $x$ is.

You start by taking derivatives on both sides, using the Chain Rule. It is important to remember that $y$ itself is a function of $x$, so when you differentiate things like $y^2$, you have to use the chain rule: $$\frac{d}{dx}y^2 = 2y\frac{dy}{dx}.$$

So, let me do that. First, we use the Chain Rule to differentiate $\sin(xy^2)$; then we will need to find the derivative of $xy^2$, which requires the Product Rule; then we will need the derivative of $y^2$, which requires the Chain Rule (as above). Let's do that: $$\begin{align*} \sin(xy^2)-x^2 &= x+5\\ \frac{d}{dx}\Bigl(\sin(xy^2)-x^2\Bigr) &= \frac{d}{dx}\Bigl(x+5\Bigr)\\ \frac{d}{dx}\sin(xy^2) - \frac{d}{dx}x^2 &= \frac{d}{dx}x + \frac{d}{dx}5\\ \cos(xy^2)\left(\frac{d}{dx}xy^2\right) - 2x &= 1\\ \cos(xy^2)\Bigl((x)'y^2 + x(y^2)'\Bigr) -2x &= 1\\ \cos(xy^2)\Bigl(y^2 + x(2yy')\Bigr) - 2x &= 1\\ y^2\cos(xy^2) + 2xy\cos(xy^2)\left(y'\right) -2x &= 1. \end{align*}$$ The next step is to "solve for $y'$". Just move every term that includes $y'$ to the left hand side, all terms that do not to the right hand side, and then divide through: $$\begin{align*} y^2\cos(xy^2) + 2xy\cos(xy^2)\left(y'\right) -2x &= 1\\ 2xy\cos(xy^2)y' &= 1 + 2x - y^2\cos(xy^2)\\ y' &= \frac{1 + 2x - y^2\cos(xy^2)}{2xy\cos(xy^2)}. \end{align*}$$ And that expresses $y'$ in terms of $x$ and $y$, given the original equation.

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    $\begingroup$ I apologize for the confusing setup. You did, in fact, infer correctly as to where I needed to go with the question. Your explanation is exactly what I needed. My problem was for some reason thinking that two solutions were required, one of the form, x = some function and also, y = some function. In any event, thanks again. $\endgroup$ – Nate222 Apr 19 '11 at 3:50
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    $\begingroup$ @NateyG: Ah, I see; you thought it said "in terms of $x$ and in terms of $y$". No, the point here is that because you do not express $y$ as an explicit function of $x$ (that is, in the form $y = $something that depends only on x), you may not be able to write $y'$ in the form $y' = $something that depends only on x. So instead you write $y'$ as a function of both $x$ and $y$, not just of $x$ as is the case with explicit functions. $\endgroup$ – Arturo Magidin Apr 19 '11 at 3:58

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