valahanyHcm

2022-11-25

I'm struggling somewhat to understand how to use implicit differentiation to solve the following equation:
$\mathrm{cos}\mathrm{cos}\left({x}^{3}{y}^{2}\right)-x\mathrm{cot}y=-2y$
I figublack 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:
$\frac{dy}{dx}\left[\mathrm{cos}\mathrm{cos}\left({x}^{3}{y}^{2}\right)-x\mathrm{cot}y\right]=\frac{dy}{dx}\left[-2y\right]$
$\frac{dy}{dx}\left[\mathrm{cos}\mathrm{cos}\left({x}^{3}{y}^{2}\right)\right]=\mathrm{sin}\mathrm{cos}\left({x}^{3}{y}^{2}\cdot {y}^{\prime }\left(x\right)\right)\cdot \mathrm{sin}\left({x}^{3}{y}^{2}\cdot {y}^{\prime }\left(x\right)\right)\cdot 6{x}^{2}y\cdot {y}^{\prime }\left(x\right)$
This seems a bit long and convoluted. I'm also not sure how this will allow me to solve for ${y}^{\prime }\left(x\right)$. Carrying on...
$\frac{dy}{dx}\left[x\mathrm{cot}y\right]=-{\mathrm{csc}}^{2}y\cdot {y}^{\prime }\left(x\right)$
$\frac{dy}{dx}\left[-2y\right]=-2$
Is my calculation correct so far? This seems to be a very complex derivative. Any comments or feedback would be appreciated.

Melanie Wong

Expert

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,
$\frac{d}{dx}\left(-2y\right)=-2\frac{dy}{dx}=-2{y}^{\prime }\left(x\right).$
On the left-hand side,
$\begin{array}{rl}\frac{d}{dx}\left[\mathrm{cos}\mathrm{cos}\left({x}^{3}{y}^{2}\right)-x\mathrm{cot}y\right]& =\frac{d}{dx}\left(\mathrm{cos}\mathrm{cos}\left({x}^{3}{y}^{2}\right)\right)-\frac{d}{dx}\left(x\mathrm{cot}y\right)\\ & =-\mathrm{sin}\mathrm{cos}\left({x}^{3}{y}^{2}\right)\cdot \frac{d}{dx}\left(\mathrm{cos}\left({x}^{3}{y}^{2}\right)\right)-\mathrm{cot}y-x\frac{d}{dx}\left(\mathrm{cot}y\right)\\ & =-\mathrm{sin}\mathrm{cos}\left({x}^{3}{y}^{2}\right)\left(-\mathrm{sin}\left({x}^{3}{y}^{2}\right)\right)\cdot \frac{d}{dx}\left({x}^{3}{y}^{2}\right)-\mathrm{cot}y+x{\mathrm{csc}}^{2}y\cdot \frac{dy}{dx}\\ & =\mathrm{sin}\mathrm{cos}\left({x}^{3}{y}^{2}\right)\mathrm{sin}\left({x}^{3}{y}^{2}\right)\left(3{x}^{2}{y}^{2}+2{x}^{3}y\frac{dy}{dx}\right)-\mathrm{cot}y+x{\mathrm{csc}}^{2}y\cdot \frac{dy}{dx}\\ & =\mathrm{sin}\mathrm{cos}\left({x}^{3}{y}^{2}\right)\mathrm{sin}\left({x}^{3}{y}^{2}\right)\left(3{x}^{2}{y}^{2}+2{x}^{3}y{y}^{\prime }\right)-\mathrm{cot}y+x{y}^{\prime }{\mathrm{csc}}^{2}y.\end{array}$
Set those two equal and solve for ${y}^{\prime }$.

Do you have a similar question?