# What is the implicit derivative of 5=x-1/(xy^2)?

Apr 18, 2018

See below.

#### Explanation:

When finding the implicit derivative of an equation, we are finding the derivative in respect to $x$.

This means that when we find the derivative of any variable other than $x$, we need to tag on a $\frac{\mathrm{ds} q u a r e}{\mathrm{dx}}$ $\left(\square \text{is the variable}\right)$ $\text{to the end} .$

Knowing this, we will start with implicit differentiation.

Here is what we are given:

$5 = x - \frac{1}{x {y}^{2}}$

We can go ahead and avoid the Quotient Rule by multiply each term by $x {y}^{2}$.

$5 = x - \frac{1}{x {y}^{2}}$

$\implies 5 x {y}^{2} = {x}^{2} {y}^{2} - 1$

Here we can use the Product Rule to solve:

$5 x {y}^{2} = {x}^{2} {y}^{2} - 1$

$\implies 5 \left({y}^{2}\right) + 5 x \left(2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 2 x \left({y}^{2}\right) + {x}^{2} \left(2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

Simplify:

$\implies 5 {y}^{2} + 10 x y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x {y}^{2} + 2 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}}$

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\implies 10 x y \frac{\mathrm{dy}}{\mathrm{dx}} - 2 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x {y}^{2} - 5 {y}^{2}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} \left(10 x y - 2 {x}^{2} y\right) = 2 x {y}^{2} - 5 {y}^{2}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x {y}^{2} - 5 {y}^{2}}{10 x y - 2 {x}^{2} y}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x y - 5 y}{10 x - 2 {x}^{2}}$

Apr 18, 2018

Please see below.

#### Explanation:

I assume that we want $\frac{\mathrm{dy}}{\mathrm{dx}}$ given that

5 = x-1/(xy^2

Method 1

To avoid the quotient rule, I would start by multiplying both sides be $x {y}^{2}$ to get

$5 x {y}^{2} = {x}^{2} {y}^{2} - 1$

Differentiating with respect to $x$ (and using the product rule) gets us

$\frac{d}{\mathrm{dx}} \left(5 x {y}^{2}\right) = \frac{d}{\mathrm{dx}} \left({x}^{2} {y}^{2} - 1\right)$

$\left(5\right) {y}^{2} + 5 x \left(2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right) = \left(2 x\right) {y}^{2} + {x}^{2} \left(2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right) - 0$

Now we'll do some algebra.

$5 {y}^{2} + 10 x y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x {y}^{2} + 2 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}}$

$10 x y \frac{\mathrm{dy}}{\mathrm{dx}} - 2 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x {y}^{2} - 5 {y}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x {y}^{2} - 5 {y}^{2}}{10 x y - 2 {x}^{2} y}$

$= \frac{2 x y - 5 y}{10 x - 2 {x}^{2}}$

Method 2

$\frac{d}{\mathrm{dx}} \left(5\right) = \frac{d}{\mathrm{dx}} \left(x - \frac{1}{x {y}^{2}}\right)$

$0 = 1 + \frac{1 {y}^{2} + 2 x y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{{x}^{2} {y}^{4}}$

$0 = 1 + \frac{y + 2 x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{{x}^{2} {y}^{3}}$

$0 = {x}^{2} {y}^{3} + y + 2 x \frac{\mathrm{dy}}{\mathrm{dx}}$

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

Resolution

To see that these are equivalent, start with

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x y - 5 y}{10 x - 2 {x}^{2}}$

$= \frac{2 x y - \left(5\right) y}{2 x \left(5\right) - 2 {x}^{2}}$

Now replace $5$ by $x - \frac{1}{x {y}^{2}}$

$= \frac{2 x y - \left(x - \frac{1}{x {y}^{2}}\right) y}{2 x \left(x - \frac{1}{x {y}^{2}}\right) - 2 {x}^{2}}$

Simplify algebraically to get

$= - \frac{{x}^{2} {y}^{3} + y}{2 x}$