Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $\cos(xy^2)(2xyy^\prime+y^2)$ – J. M. Apr 18 '11 at 21:54
Look up the Implicit Function Theorem. – Yuval Filmus Apr 18 '11 at 22:00
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. – Graphth Apr 18 '11 at 22:08
up vote 5 down vote accepted

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.

share|cite|improve this answer
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. – Nate222 Apr 19 '11 at 3:50
@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. – Arturo Magidin Apr 19 '11 at 3:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.