# How do you implicitly differentiate  x^2+x/y-xy^2+x=3y ?

Jun 30, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x {y}^{2} + y - {y}^{4} + {y}^{2}}{3 {y}^{2} + x + 2 x {y}^{3}} = {f}^{'} \left(x\right)$

#### Explanation:

$\textcolor{b l u e}{\text{The teaching bit}}$
By example: Suppose we had ${y}^{2} + 3 y = {x}^{4}$

Let $z = {y}^{2} + 3 y \to \frac{\mathrm{dz}}{\mathrm{dy}} = 2 y + 3$...................(1)
Then $z = {x}^{4} \to \frac{\mathrm{dz}}{\mathrm{dx}} = 4 {x}^{3}$...............................(2)

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Consider equation (1). If we can convert $\frac{\mathrm{dz}}{\mathrm{dy}}$ to $\frac{\mathrm{dz}}{\mathrm{dx}}$ then we can directly equate the Eqn(1) to Eqn(2)

We use:$\text{ } \frac{\mathrm{dz}}{\cancel{\mathrm{dy}}} \times \frac{\cancel{\mathrm{dy}}}{\mathrm{dx}}$

But from Eqn(1)

$\frac{\mathrm{dz}}{\mathrm{dy}} = 2 y + 3 \text{ }$ so $\text{ "(dz)/(dy)xx(dy)/(dx)" "=" } \frac{\mathrm{dy}}{\mathrm{dx}} \left(2 y + 3\right)$

so we have $\frac{\mathrm{dy}}{\mathrm{dx}} \left(2 y + 2\right) = \frac{\mathrm{dz}}{\mathrm{dx}} = 4 {x}^{3}$

Thus $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 {x}^{3}}{2 y + 2}$

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$\textcolor{b l u e}{\text{Answering the question}}$
Differentiating each term independently

As appropriate use $\frac{d}{\mathrm{dx}} \left(u v\right) = v \frac{\mathrm{du}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}}$

As appropriate use $\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$
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$\frac{d}{\mathrm{dx}} \left({x}^{2}\right) = \textcolor{b r o w n}{+ 2 x}$

$\frac{d}{\mathrm{dx}} \left(\frac{x}{y}\right) = \textcolor{b r o w n}{\frac{y - x \frac{\mathrm{dy}}{\mathrm{dx}}}{y} ^ 2}$

d/(dx)(-xy^2) = -[y^2+x2y(dy)/(dx)]=color(brown)(-y^2-2xy(dy)/(dx)

$\frac{d}{\mathrm{dx}} \left(x\right) = \textcolor{b r o w n}{+ 1}$

$\frac{d}{\mathrm{dx}} \left(3 y\right) = \textcolor{b r o w n}{3 \frac{\mathrm{dy}}{\mathrm{dx}}}$
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$\textcolor{b r o w n}{\text{Putting it all together}}$

The differentiated equation is:

$2 x + \frac{y - x \frac{\mathrm{dy}}{\mathrm{dx}}}{y} ^ 2 - {y}^{2} - 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} + 1 = 3 \frac{\mathrm{dy}}{\mathrm{dx}}$

Collecting like terms

$3 \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{x \frac{\mathrm{dy}}{\mathrm{dx}}}{y} ^ 2 + 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x + \frac{\cancel{y}}{y} ^ \left(\cancel{2}\right) - {y}^{2} + 1$

$\frac{3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + x \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}}}{y} ^ 2 = 2 x + \frac{1}{y} - {y}^{2} + 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(3 {y}^{2} + x + 2 x {y}^{3}\right) = {y}^{2} \left(2 x + \frac{1}{y} - {y}^{2} + 1\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x {y}^{2} + y - {y}^{4} + {y}^{2}}{3 {y}^{2} + x + 2 x {y}^{3}}$