valahanyHcm

Answered

2022-11-25

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

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

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:

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

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

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

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

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

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

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

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:

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

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

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

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

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

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

Answer & Explanation

Melanie Wong

Expert

2022-11-26Added 11 answers

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,

$\frac{d}{dx}(-2y)=-2\frac{dy}{dx}=-2{y}^{\prime}(x).$

On the left-hand side,

$\begin{array}{rl}\frac{d}{dx}[\mathrm{cos}\mathrm{cos}({x}^{3}{y}^{2})-x\mathrm{cot}y]& =\frac{d}{dx}(\mathrm{cos}\mathrm{cos}({x}^{3}{y}^{2}))-\frac{d}{dx}(x\mathrm{cot}y)\\ & =-\mathrm{sin}\mathrm{cos}({x}^{3}{y}^{2})\cdot \frac{d}{dx}(\mathrm{cos}({x}^{3}{y}^{2}))-\mathrm{cot}y-x\frac{d}{dx}(\mathrm{cot}y)\\ & =-\mathrm{sin}\mathrm{cos}({x}^{3}{y}^{2})(-\mathrm{sin}({x}^{3}{y}^{2}))\cdot \frac{d}{dx}({x}^{3}{y}^{2})-\mathrm{cot}y+x{\mathrm{csc}}^{2}y\cdot \frac{dy}{dx}\\ & =\mathrm{sin}\mathrm{cos}({x}^{3}{y}^{2})\mathrm{sin}({x}^{3}{y}^{2})(3{x}^{2}{y}^{2}+2{x}^{3}y\frac{dy}{dx})-\mathrm{cot}y+x{\mathrm{csc}}^{2}y\cdot \frac{dy}{dx}\\ & =\mathrm{sin}\mathrm{cos}({x}^{3}{y}^{2})\mathrm{sin}({x}^{3}{y}^{2})(3{x}^{2}{y}^{2}+2{x}^{3}y{y}^{\prime})-\mathrm{cot}y+x{y}^{\prime}{\mathrm{csc}}^{2}y.\end{array}$

Set those two equal and solve for ${y}^{\prime}$.

$\frac{d}{dx}(-2y)=-2\frac{dy}{dx}=-2{y}^{\prime}(x).$

On the left-hand side,

$\begin{array}{rl}\frac{d}{dx}[\mathrm{cos}\mathrm{cos}({x}^{3}{y}^{2})-x\mathrm{cot}y]& =\frac{d}{dx}(\mathrm{cos}\mathrm{cos}({x}^{3}{y}^{2}))-\frac{d}{dx}(x\mathrm{cot}y)\\ & =-\mathrm{sin}\mathrm{cos}({x}^{3}{y}^{2})\cdot \frac{d}{dx}(\mathrm{cos}({x}^{3}{y}^{2}))-\mathrm{cot}y-x\frac{d}{dx}(\mathrm{cot}y)\\ & =-\mathrm{sin}\mathrm{cos}({x}^{3}{y}^{2})(-\mathrm{sin}({x}^{3}{y}^{2}))\cdot \frac{d}{dx}({x}^{3}{y}^{2})-\mathrm{cot}y+x{\mathrm{csc}}^{2}y\cdot \frac{dy}{dx}\\ & =\mathrm{sin}\mathrm{cos}({x}^{3}{y}^{2})\mathrm{sin}({x}^{3}{y}^{2})(3{x}^{2}{y}^{2}+2{x}^{3}y\frac{dy}{dx})-\mathrm{cot}y+x{\mathrm{csc}}^{2}y\cdot \frac{dy}{dx}\\ & =\mathrm{sin}\mathrm{cos}({x}^{3}{y}^{2})\mathrm{sin}({x}^{3}{y}^{2})(3{x}^{2}{y}^{2}+2{x}^{3}y{y}^{\prime})-\mathrm{cot}y+x{y}^{\prime}{\mathrm{csc}}^{2}y.\end{array}$

Set those two equal and solve for ${y}^{\prime}$.

Most Popular Questions