# Implicit Differentiation

In this section we learn how to perform implicit differentiation.
This technique will allow us to differentiate expressions in which $$y$$ is not explicitly defined as a function of $$x$$.
For instance, expressions like: $3 y^2x - sin(y)+4x^2=3$ The method, or technique for differentiating expressions in which $$y$$ is implicitly defined is shown here.

## Method - Implicit Differentiation

To differentiate functions in which $$y$$ is implicitly defined, two things should be kept in mind:

• $$y$$ is a function of $$x$$.
In fact when we start learning implicit differentiation it's often useful to replace every $$y$$ we see by $$y(x)$$.

• The chain rule: $$\begin{pmatrix} g\begin{bmatrix} h(x)\end{bmatrix}\end{pmatrix}' = h'(x)\times g'\begin{pmatrix}h(x) \end{pmatrix}$$.
Consequently, to differentiate composite functions involing $$y$$, we use the chain rule: $\begin{pmatrix}g\begin{bmatrix}y(x)\end{bmatrix}\end{pmatrix}' = y'(x)\times g'\begin{bmatrix}y(x)\end{bmatrix}$ We'll usually write this as: $\begin{pmatrix}g\begin{bmatrix}y\end{bmatrix}\end{pmatrix}' = \frac{dy}{dx}\times g'\begin{bmatrix}y\end{bmatrix}$

## Example

Differentiate the expression that follows with respect to $$x$$: $3y^2-x^2=8$

### Solution

We can re-write $$3y^2-x^2=8$$ highlighting the fact that $$y$$ is a function of $$x$$, that's: $3y(x)^2-x^2=8$ The key when differentiating this is to treat the expression $$y(x)^2$$ in exactly the same way that we would treat composite expressions like $$\begin{pmatrix}sin(x)\end{pmatrix}^2$$, or $$\begin{pmatrix}x^2-1\end{pmatrix}^2$$, ... .

Using the chain rule, we find that the derivative of $$y(x)^2$$ is $$2y'(x).y(x)$$, that's: $\begin{pmatrix}y(x)^2\end{pmatrix}' = 2y'(x)y(x)$
We tend to write this in the form: $\begin{pmatrix}y^2\end{pmatrix}' = 2\frac{dy}{dx}y$

We differentiate the remaining terms of the expression the usual way.
So differentiating both sides of $$3y^2-x^2=8$$ with respect to $$x$$: $3\times 2\frac{dy}{dx}y - 2x = 0$ That's: $6\frac{dy}{dx}y - 2x = 0$ And that would be the final answer.

We could also go further and find an expression for $$\frac{dy}{dx}$$. To do so we rearrange the result we just found, to make $$\frac{dy}{dx}$$ the subject: $6\frac{dy}{dx}y = 2x$ That's: $\frac{dy}{dx} = \frac{2x}{6y}$ Finally: $\frac{dy}{dx} = \frac{x}{3y}$

## Exercise 1

1. Given $$y^2-x=4$$, find an expression for $$\frac{dy}{dx}$$.
2. Given $$x^3+y^3 = 5$$, find an expression for $$\frac{dy}{dx}$$.
3. Given $$y=2x.sin(y)$$, find an expression for $$\frac{dy}{dx}$$.
4. Given $$4x.\sqrt{sin(y)}-5y^2=1$$, find an expression for $$\frac{dy}{dx}$$.

1. Given $$y^2-x=4$$, we find: $\frac{dy}{dx} = \frac{1}{2y}$
2. Given $$x^3+y^3 = 5$$, we find: $\frac{dy}{dx} = -\frac{x^2}{y^2}$
3. Given $$y=2x.sin(y)$$, we find: $\frac{dy}{dx} = \frac{2.sin(y)}{1-2x.cos(y)}$
4. Given $$4x.\sqrt{sin(y)}-5y^2=1$$, we find: $\frac{dy}{dx} = \frac{2.sin(y)}{5y.\sqrt{sin(y)}-x.cos(y)}$

## Second and Higher Derivatives

At times we'll be required to find an expression for the second derivative $$\frac{d^2y}{dx^2}$$ (or higher).

To do this we "simply" differentiate the expression, with respect to $$x$$, twice and rearrange to make $$\frac{d^2y}{dx^2}$$ the subject.

## Example

Given the expression: $3x^2+y^2 = 7$

1. $$\frac{dy}{dx}$$
2. $$\frac{d^2y}{dx^2}$$

### Solution

We start by differentiating with respect to $$x$$: $3x^2+y^2 = 7$ becomes: $6x+2\frac{dy}{dx}y = 0$

1. To find $$\frac{dy}{dx}$$ we rearrange this expression to make $$\frac{dy}{dx}$$ the subject, that's: $2\frac{dy}{dx}y = -6x$ dividing both sides by $$2y$$ we obtain: $\frac{dy}{dx} = -\frac{3x}{y}$
2. To find $$\frac{d^2y}{dx^2}$$ we differentiate the first differentiated expression we found $6x+2\frac{dy}{dx}y = 0$ with respect to $$x$$ again. Keep in mind that to differentiate the term $$2\frac{dy}{dx}y$$ we use the product rule, leading to: $6+2\frac{d^2y}{dx^2}y+2\frac{dy}{dx}\frac{dy}{dx} = 0$ which can be written: $6+2\frac{d^2y}{dx^2}y+2\begin{pmatrix}\frac{dy}{dx}\end{pmatrix}^2 = 0$ We can divide throughout by $$2$$ to write: $3+\frac{d^2y}{dx^2}y+\begin{pmatrix}\frac{dy}{dx}\end{pmatrix}^2 = 0$ We now rearrange this to make $$\frac{d^2y}{dx^2}$$ the subject: $\frac{d^2y}{dx^2}y = -3 -\begin{pmatrix}\frac{dy}{dx}\end{pmatrix}^2$ Finally, dividing throughout by $$y$$: $\frac{d^2y}{dx^2} = \frac{-3 -\begin{pmatrix}\frac{dy}{dx}\end{pmatrix}^2}{y}$ That's: $\frac{d^2y}{dx^2} = -\frac{3 +\begin{pmatrix}\frac{dy}{dx}\end{pmatrix}^2}{y}$

## Exercise 2

1. Given the expression $$sin(y)-cos(x)=1$$, find an expression for:
1. $$\frac{dy}{dx}$$
2. $$\frac{d^2y}{dx^2}$$
2. Given the expression $$3x^2y-cos(y)=4$$, find an expression for:
1. $$\frac{dy}{dx}$$
2. $$\frac{d^2y}{dx^2}$$

1. Given the expression $$sin(y)-cos(x)=1$$, we find:
1. $$\frac{dy}{dx} = -\frac{sin(x)}{cos(y)}$$
2. $$\frac{d^2y}{dx^2} = \begin{pmatrix}\frac{dy}{dx} \end{pmatrix}^2.tan(y)-\frac{cos(x)}{cos(y)}$$
2. Given the expression $$3x^2y-cos(y)=4$$, we find:
1. $$\frac{dy}{dx} = -\frac{6xy}{3x^2+sin(y)}$$
2. $$\frac{d^2y}{dx^2} = - \frac{6y+12x.\frac{dy}{dx} + \begin{pmatrix}\frac{dy}{dx}\end{pmatrix}^2.cos(y)}{3x^2+sin(y)}$$