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.
Best Answer
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 $y'$.