# How do you implicitly differentiate csc(x^2/y^2)=e^-x-y ?

##### 1 Answer
Mar 12, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{y}^{3} {e}^{-} x - 2 x y \csc \left(\frac{{x}^{2}}{{y}^{2}}\right) \cot \left(\frac{{x}^{2}}{{y}^{2}}\right)}{- {y}^{3} - 2 {x}^{2} \csc \left(\frac{{x}^{2}}{{y}^{2}}\right) \cot \left(\frac{{x}^{2}}{{y}^{2}}\right)}$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left(\csc \left(\frac{{x}^{2}}{{y}^{2}}\right) = {e}^{-} x - y\right)$
First we find the derivative of the left side of the function d/dx (csc((x^2)/(y^2)) This is a composition of functions. Meaning we apply the Chain Rule. where $f \left(x\right) = \csc \left(x\right)$ and $g \left(x\right) = \frac{{x}^{2}}{{y}^{2}}$ where $f \left(g \left(x\right)\right) = \csc \left(\frac{{x}^{2}}{{y}^{2}}\right)$

$\frac{d}{\mathrm{dx}} f \left(x\right) = \csc \left(x\right) = - \csc \left(x\right) \cot \left(x\right)$ we must substitute of $x = g \left(x\right)$ , so $\frac{d}{\mathrm{dx}} \csc \left(g \left(x\right)\right) = - \csc \left(\frac{{x}^{2}}{{y}^{2}}\right) \cot \left(\frac{{x}^{2}}{{y}^{2}}\right)$

$\frac{d}{\mathrm{dx}} g \left(x\right) = \frac{2 x {y}^{2} - 2 y {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}}{{y}^{4}}$ apply the quotient rule $\frac{f ' g - f g '}{g} ^ 2$ where $f = {x}^{2}$ and $g = {y}^{2}$

Furthermore, $\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = \left(- \csc \left(\frac{{x}^{2}}{{y}^{2}}\right) \cot \left(\frac{{x}^{2}}{{y}^{2}}\right)\right) \frac{2 x {y}^{2} - 2 y {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}}{{y}^{4}}$ let us simplify this function now as to make our lives easier in the following minutes.
We multiple $\left(- \csc \left(\frac{{x}^{2}}{{y}^{2}}\right) \cot \left(\frac{{x}^{2}}{{y}^{2}}\right)\right)$ by both term $2 x {y}^{2}$ and term $- 2 y {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}$ resulting in a function similarly looking to that of

$\frac{2 y {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} \csc \left(\frac{{x}^{2}}{{y}^{2}}\right) \cot \left(\frac{{x}^{2}}{{y}^{2}}\right) - 2 x {y}^{2} \csc \left(\frac{{x}^{2}}{{y}^{2}}\right) \cot \left(\frac{{x}^{2}}{{y}^{2}}\right)}{{y}^{4}}$ we see that the top function has a like term of $y$ , so
$\frac{2 {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} \csc \left(\frac{{x}^{2}}{{y}^{2}}\right) \cot \left(\frac{{x}^{2}}{{y}^{2}}\right) - 2 x y \csc \left(\frac{{x}^{2}}{{y}^{2}}\right) \cot \left(\frac{{x}^{2}}{{y}^{2}}\right)}{{y}^{3}}$

We are not finished we must now find the derivative of the right side of the function. $\frac{d}{\mathrm{dx}} {e}^{-} x - y$ substituting $u = - x$ $\frac{d}{\mathrm{dx}} {e}^{u} = {e}^{-} x$
$\frac{d}{\mathrm{dx}} - x = - 1$
hence $\frac{d}{\mathrm{dx}} {e}^{u} = - {e}^{-} x$
$\frac{d}{\mathrm{dx}} - y = - \frac{\mathrm{dy}}{\mathrm{dx}}$

Therefore $\frac{d}{\mathrm{dx}} {e}^{-} x - y = - {e}^{-} x - \frac{\mathrm{dy}}{\mathrm{dx}}$

Furthermore,  (2x^2dy/dxcsc((x^2)/(y^2))cot((x^2)/(y^2))-2xycsc((x^2)/(y^2))cot((x^2)/(y^2)))/(y^3) = -e^-x-dy/dx

From here we must move $\frac{\mathrm{dy}}{\mathrm{dx}}$ to one side of our function.
Meaning first multiple the left side of the function by ${y}^{3}$ moving this term to the right. resulting in,
$\left(2 {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} \csc \left(\frac{{x}^{2}}{{y}^{2}}\right) \cot \left(\frac{{x}^{2}}{{y}^{2}}\right) - 2 x y \csc \left(\frac{{x}^{2}}{{y}^{2}}\right) \cot \left(\frac{{x}^{2}}{{y}^{2}}\right)\right) = - {y}^{3} {e}^{-} x - {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}}$

From here let us subtract $2 {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} \csc \left(\frac{{x}^{2}}{{y}^{2}}\right) \cot \left(\frac{{x}^{2}}{{y}^{2}}\right)$ from both sides because it has the term $\frac{\mathrm{dy}}{\mathrm{dx}}$ resulting in,
$- 2 x y \csc \left(\frac{{x}^{2}}{{y}^{2}}\right) \cot \left(\frac{{x}^{2}}{{y}^{2}}\right) = - {y}^{3} {e}^{-} x - {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} - 2 {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} \csc \left(\frac{{x}^{2}}{{y}^{2}}\right) \cot \left(\frac{{x}^{2}}{{y}^{2}}\right)$

Now let us remove any terms on the right side that aren't $\frac{\mathrm{dy}}{\mathrm{dx}}$ , so
${y}^{3} {e}^{-} x - 2 x y \csc \left(\frac{{x}^{2}}{{y}^{2}}\right) \cot \left(\frac{{x}^{2}}{{y}^{2}}\right) = - {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} - 2 {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} \csc \left(\frac{{x}^{2}}{{y}^{2}}\right) \cot \left(\frac{{x}^{2}}{{y}^{2}}\right)$ From here factor out $\frac{\mathrm{dy}}{\mathrm{dx}}$ from the right of our equation ${y}^{3} {e}^{-} x - 2 x y \csc \left(\frac{{x}^{2}}{{y}^{2}}\right) \cot \left(\frac{{x}^{2}}{{y}^{2}}\right) = \frac{\mathrm{dy}}{\mathrm{dx}} \left(- {y}^{3} - 2 {x}^{2} \csc \left(\frac{{x}^{2}}{{y}^{2}}\right) \cot \left(\frac{{x}^{2}}{{y}^{2}}\right)\right)$ now divide both sides by our factored expression resulting in our final answer.
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{y}^{3} {e}^{-} x - 2 x y \csc \left(\frac{{x}^{2}}{{y}^{2}}\right) \cot \left(\frac{{x}^{2}}{{y}^{2}}\right)}{- {y}^{3} - 2 {x}^{2} \csc \left(\frac{{x}^{2}}{{y}^{2}}\right) \cot \left(\frac{{x}^{2}}{{y}^{2}}\right)}$