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I'm struggling somewhat to understand how to use implicit differentiation to solve the following equation:

$$\cos\cos(x^3y^2) - x \cot y = -2y$$

I figured that the calculation requires the chain rule to differentiate the composite function, but I'm not sure how to 'remove' the y with respect to x from inside the composite function. My calculations are:

$$\frac{dy}{dx}[\cos\cos(x^3y^2) - x \cot y] = \frac{dy}{dx}[-2y]$$

$$\frac{dy}{dx}[\cos\cos(x^3y^2)] = \sin \cos (x^3y^2 \cdot y'(x)) \cdot \sin (x^3y^2 \cdot y'(x)) \cdot 6x^2y\cdot y'(x)$$

This seems a bit long and convoluted. I'm also not sure how this will allow me to solve for $y'(x)$. Carrying on...

$$\frac{dy}{dx}[x \cot y] = -\csc^2y \cdot y'(x)$$

$$\frac{dy}{dx}[-2y] = -2$$

Is my calculation correct so far? This seems to be a very complex derivative. Any comments or feedback would be appreciated.

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First, you should be writing $\frac{d}{dx}$, not $\frac{dy}{dx}$. $\frac{dy}{dx}$ refers to the derivative of $y$ with respect to $x$, while here you are taking the derivative of some complicated function with respect to $x$. After that, this is just an application of the chain rule. On the right-hand side, $$\frac{d}{dx}(-2y) = -2\frac{dy}{dx} = -2y'(x).$$ On the left-hand side, \begin{align} \frac{d}{dx}[\cos\cos(x^3y^2) - x \cot y] &= \frac{d}{dx}(\cos\cos(x^3y^2)) - \frac{d}{dx}(x\cot y) \\ &= -\sin\cos(x^3y^2)\cdot\frac{d}{dx}(\cos(x^3y^2)) - \cot y - x\frac{d}{dx}(\cot y) \\ &= -\sin\cos(x^3y^2)\left(-\sin(x^3y^2)\right)\cdot\frac{d}{dx}(x^3y^2) - \cot y + x\csc^2 y\cdot\frac{dy}{dx} \\ &= \sin\cos(x^3y^2)\sin(x^3y^2)\left(3x^2y^2 + 2x^3y\frac{dy}{dx}\right) - \cot y + x\csc^2 y\cdot\frac{dy}{dx} \\ &= \sin\cos(x^3y^2)\sin(x^3y^2)(3x^2y^2 + 2x^3yy') - \cot y + xy'\csc^2 y. \end{align} Set those two equal and solve for $y'$.

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    $\begingroup$ I think you want to square the $\csc y$ in the last terms of the last 3 expressions. $\endgroup$ – user84413 Aug 25 '14 at 19:43
  • $\begingroup$ @user84413 thanks. $\endgroup$ – rogerl Aug 25 '14 at 22:03
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$$\frac{d}{dx}(\cos\cos(x^3y^2)) = -\sin(\cos(x^3y^2)\cdot(- \sin(x^3y^2))\cdot(3x^2y^2 + 2x^3yy') = \\ \sin(\cos(x^3y^2)\cdot(\sin(x^3y^2))\cdot(3x^2y^2 + 2x^3yy')$$

$$\frac{d}{dx} (x\cot(y)) = \cot(y) - x\csc^2(y)(y')$$

$$\frac{d}{dx}(-2y) = -2y'$$

Can you take it from here (isolating $y' = \dfrac{dy}{dx}$)?

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  • $\begingroup$ Ah, so the product rule is required for the innermost term? $\endgroup$ – thisisanon Aug 25 '14 at 19:02
  • $\begingroup$ Yes, it is required. $\endgroup$ – Namaste Aug 25 '14 at 19:02
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    $\begingroup$ Also, shouldn't $\frac{d}{dx}(\cot y) = -\csc^2y$? $\endgroup$ – thisisanon Aug 25 '14 at 19:32
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$$\cos\cos(x^3y^2) - x \cot y = -2y$$ $$-\sin(\cos(x^3y^2))(-\sin(x^3y^2)(3x^2y^2+2x^2yy')-\cot y-x(\frac{\cos y}{siny})= -2y'$$ $$\sin(\cos(x^3y^2))(\sin(x^3y^2)(3x^2y^2+2x^2yy')-\cot y+\frac{xy'}{\sin^2y}= -2y'$$

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  • $\begingroup$ Your coefficients of $12$ should be a $2$. $\endgroup$ – Namaste Aug 25 '14 at 19:17
  • $\begingroup$ Yes. Thank you. Corrected. $\endgroup$ – Adi Dani Aug 25 '14 at 19:20

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