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For an assignment, I've been asked to find the $\frac{dy}{dx}$ for the formula $$2 \cos(4x) \sin(9y)=7$$

The major problem I'm having with understanding implicit differentiation is understanding what distinguishes a typical $\frac{d}{dx}$ from $\frac{dy}{dx}$ An example I came across gave the process of $$\frac{d}{dx}[xy^{2}]= x\frac{d}{dx}[y^{2}]+y^{2}\frac{d}{dx}[x]= x(2y\frac{dy}{dx}+y^{2}(1)= 2xy\frac{dy}{dx}+y^{2}$$

Applying that approach to the above formula, I came up with

$$2 \cos(4x) \sin (9y)=7 \rightarrow \frac{d}{dx}[2 \cos(4x) \sin(9y)]= \frac{d}{dx}(7)$$

Applying the product rule to the elements:

$$2[\cos(4x) \frac{d}{dx}\sin(9y)]+[\sin(9y)\frac{d}{dx} \cos(4x)]=0$$ $$2[\cos(4x)9 \cos(9y)\frac{dy}{dx}]+[\sin(9y)(-4 \sin(4x))\frac{dy}{dx}]=0$$ $$18 \cos(4x)\cos(9y)\frac{dy}{dx} + \sin(9y)-4 \sin(4x)\frac{dy}{dx}=0$$

First, am I on the right track as far as assigning the correct $\frac{dy}{dx}$ terms? Assuming I am, how do I separate the $\frac{dy}{dx}$ terms from their respective functions? Because if I divide $18 \cos(4x)$ to isolate $\sin(9y)$, dividing against zero will produce zero.

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I'd like to mention that $\frac{dy}{dx}$ can be treated just like any other term. So, if you did end up with two terms containing the derivative, it can be factored out like: $$a \frac{dy}{dx} + b \frac{dy}{dx} = (a+b)\frac{dy}{dx}$$. You just have to remember that for the most part $\frac{dy}{dx}$ is a single term, like $\cos(x)$ is. –  rcollyer Mar 9 '11 at 3:52

2 Answers 2

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Going from $\frac{d}{dx}[2 \cos(4x) \sin(9y)]=0$ to $2[\cos(4x) \frac{d}{dx}\sin(9y)]+[\sin(9y)\frac{d}{dx} \cos(4x)]=0$ you lost a factor $2$ on the second term. Then, considering only the second term $[\sin(9y)\frac{d}{dx} \cos(4x)]=[\sin(9y)(-4\sin(4x)]$ there should not be a $\frac{dy}{dx}$ as the argument of the $\cos$ does not depend on $y$, just on $x$. From your next to last line to the last line you lost some parentheses that subtracted the terms instead of multiplying them.

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If I understand you right, should the final answer be $$-\frac{18 \cos(4x)\cos(9y)}{8 \sin(4x)\sin(9y)}$$? This is assuming that functions relying on dissimilar variables can't be combined. –  Jason Mar 8 '11 at 20:11
    
This is inverted and the sign is wrong. From $18 \cos(4x)\cos(9y)\frac{dy}{dx} - \sin(9y)4 \sin(4x)=0$ you take the second term across and there is no minus sign, then divide by $18\cos(4x)\cos(9y)$ (assuming it is non-zero) –  Ross Millikan Mar 8 '11 at 20:15
    
Understood. In previous examples, the $\frac{dy}{dx}$ was always on the second term, and that tripped me up this time. Thanks for the help! –  Jason Mar 8 '11 at 20:23

Two mistakes: you lost a factor of $2$, and you introduced a spurious $\frac{dy}{dx}$.

$\frac{dy}{dx}$ only comes in when you are differentiating a function of $y$ (or $y$ itself), because $y$ is itself a function of $x$, and so the Chain Rule applies; you don't know what the derivative of $y$ is, so you leave it indicated with $\frac{dy}{dx}$.

It might be less confusing if you use $y'$ instead of $y$ in this instance.

So, we have: \begin{align*} 2\cos(4x)\sin(9y) &= 7\\ \frac{d}{dx}\Bigl(2\cos(4x)\sin(9y)\Bigr) &= \frac{d}{dx}7\\ 2\frac{d}{dx}\Bigl(\cos(4x)\sin(9y)\Bigr) &= 0\quad\mbox{(factor out a $2$)}\\ 2\Biggl( \cos(4x)\Bigl(\frac{d}{dx}\sin(9y)\Bigr) + \sin(9y)\Bigl(\frac{d}{dx}\cos(4x)\Bigr)\Biggr) &= 0\quad\mbox{(product rule)}\\ 2\Biggl( \cos(4x)\Bigl( \cos(9y)\frac{d}{dx}(9y)\Bigr) + \sin(9y)\Bigl(-\sin(4x)\frac{d}{dx}(4x)\Bigr)\Biggr) &=0\quad\mbox{(chain rule)}\\ 2\Biggl( \cos(4x)\Bigl(\cos(9y)9\frac{d}{dx}(y)\Bigr) + \sin(9y)\Bigl(-\sin(4x)4\Bigr)\Biggr) &= 0 \quad\mbox{(chain rule)}\\ 2\Biggl(\cos(4x)\Bigl(9\cos(9y)y'\Bigr) + \sin(9y)\Bigl(-4\sin(4x)\Bigr)\Biggr) &= 0 \quad\mbox{(because $\frac{d}{dx}y = y'$)}\\ 18\cos(4x)\cos(9y)y' - 8\sin(9y)\sin(4x) &= 0\quad\mbox{(algebra)}\\ 18\cos(4x)\cos(9y)y' &= 8\sin(9y)\sin(4x)\\ y' &= \frac{8\sin(9y)\sin(4x)}{18\cos(4x)\cos(9y)}\\ \frac{dy}{dx} &= \frac{4}{9}\left(\frac{\sin(9y)}{\cos(9y)}\right)\left(\frac{\sin(4x)}{\cos(4x)}\right)\\ \frac{dy}{dx} &= \frac{4}{9}\tan(9y)\tan(4x). \end{align*}

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