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Find $\frac{dy}{dx}$ given that $\sin(8x+y)=18x$.

So far, I've done the chain rule and find the derivatives for both sides so I got:

$\cos(8x+y)(8+\frac{dy}{dx})=18$

From here I'm lost, do I subtract 18 from both sides and then multiply it by 8 so I get:

$-144\cos(8x+y)(\frac{dy}{dx})=0$ ?

Please Help!!!

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  • $\begingroup$ The derivative is wrong in the third line. It should be $\cos(8x+y) (8+\frac{dy}{dx})$ $\endgroup$ – copper.hat Sep 29 '13 at 20:17
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    $\begingroup$ Hint: If you have a linear equation $a\times (b+cx)=d$ how do you solve for $x$? $\endgroup$ – Mark Bennet Sep 29 '13 at 20:17
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    $\begingroup$ If you want to express in terms of $x$ only, use $\cos(8x+y)=\pm\sqrt{1-(18x)^2}$. $\endgroup$ – André Nicolas Sep 29 '13 at 20:19
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Your problems are with algebraic manipulation. You have a linear equation to solve for $\frac{dy}{dx}$. If you were trying to solve $3(5+T)=7$ for $T$, you could divide by $3$ to get $5+T=\frac{7}{3}$, then subtract $5$ to get $T=\frac{7}{3}-5$ (which you might rewrite as $T=-\frac{8}{3}$). Or, you could distribute the multiplication on the left first, to get $15+3T=7$, then subtract $15$ to get $3T=-8$, then divide by $3$ to get $T=-\frac{8}{3}$.

Your equation has the same general form, in that $\dfrac{dy}{dx}$ plays the role of $T$ above, and the other parts can be thought of as numbers to subtract, add, muliply, or divide according to correct algebraic rules.

From here I'm lost, do I subtract 18 from both sides and then multiply it by 8 so I get:

$-144\cos(8x+y)(\frac{dy}{dx})=0$ ?

Here you are saying that you got $\cos(8x+y)(8+\frac{dy}{dx})-18=-144\cos(8x+y)(\frac{dy}{dx})$, and that would be like saying $3(5+T)-7 = -35\cdot3(T)$; it does not follow from principles of arithmetic that numbers can be rearranged in such ways. You can use principles like $(ab)/a = b$, $(a+b)-a = b$, and $a(b+c)= ab+ac$.

You want to isolate $\dfrac{dy}{dx}$; in order to do so, you can divide by $\cos(8x+y)$ to get $8+\frac{dy}{dx}=\dfrac{18}{\cos(8x+y)}$. Then you can subtract $8$ to get $\frac{dy}{dx}=\dfrac{18}{\cos(8x+y)}-8$. André Nicolas points out in a comment how you can solve entirely in terms of $x$, similar to Pocho la pantera's answer, except that the sign of $\cos(8x+y)$ is not clear, hence which root to take is not clear.

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Hint: $$y=\arcsin (18x) - 8x.$$

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    $\begingroup$ Surely the implicit function approach would be easier? $\endgroup$ – copper.hat Sep 29 '13 at 20:18
  • $\begingroup$ The implicit function approach is also more correct. The equation in this answer does not follow unless $8x+y$ is restricted to the interval $[-\pi/2,\pi/2]$. $\endgroup$ – Jonas Meyer Sep 29 '13 at 20:19
  • $\begingroup$ @copper.hat would I have to add cos to both sides so because the derivative has to be on the opposite side? $\endgroup$ – John Beal Sep 29 '13 at 20:27
  • $\begingroup$ @JohnBeal: Why would you add $\cos$? If you were trying to solve $3(5+T)=7$ for $T$, you would not add $3$. Mark Bennet's hint is a good one. $\endgroup$ – Jonas Meyer Sep 29 '13 at 20:36
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$$ \cos(8x+y)(8+\frac{dy}{dx})=18 $$ For the other side $$ \cos(8x+y)=\sqrt{1-\sin^2(8x+y)}=\sqrt{1-(18x)^2} $$ Then $$ \frac{dy}{dx}=\frac{18}{\sqrt{1-(18x)^2}}-8 $$

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    $\begingroup$ This is true if an assumption is added that $\cos(8x+y)> 0$. This answer requires that $8x+y$ is restricted to intervals of the form $(-\pi/2+2k\pi,\pi/2+2k\pi)$, $k\in\mathbb Z$. $\endgroup$ – Jonas Meyer Sep 29 '13 at 20:58
  • $\begingroup$ Yes, you are right. Otherwise we have to take negative square root. $\endgroup$ – Pocho la pantera Sep 29 '13 at 21:13

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