# Implicit Differentiation

## Implicit Differentiation

Previously, we have been completely working with explicit differentiation. This is where we can write an equation with one term purely on itself equalling something else. Such as y = f(x) or a = 3b + 2c. Even when it is written as something like y2 - x 2 = 3x + 1, we can write it explicitly like, y = (x2 + 3x + 1).

There are times when we can not write our terms explicitly, like,

x - cos(x) = y4 + y + 3.

This does not mean we can not find the derivative of it. We do this with implicit differentiation. Generally, in implicit differentiation, it is expected and assumed that we take y as a function of x.

### Definition - Implicit Differentiation

To calculate a derivative of y, with implicit differentiation, differentiate each side of the equation and any time we differentiate y in the equation, treat it as a function of x. It might seem confusing but the examples will help explain.

In the same way we differentiate an equation with multiple terms by differentiating the individual terms of an equation like this,

y = 3x2 + 2x + 1

(y) = (3x2) + (2x) + (1)

= 6x + 2 + 0

We apply the same method when we have a y term that is on “the other side” of the equation,

Mathematically speaking the equation below is not true, but say strictly for explanations sake we have,

y = x + y + 1

Then, using Implicit Differentiation,

D(y) = D(x) + D(y) + D(1)

y’ = 1 + + 0

We are symbolizing the derivative of y as dy/dx, because in implicit differentiation, we take the derivative of y to be a derivative of y as a function of x. It is as if we have another y equation within our y equation.

### Rules to Remember

For some of these exercises we will recall some of our derivative equations,

Power Rule for functions - D( f(x)^n ) = n( f(x)^{n-1} )( f’((x) )

Product Rule - D( f(x)g(x) ) = f’(x)g(x) + f(x)g’(x)

Chain Rule - D( f( g(x) ) ) = f’( g(x) )g’(x)

### Beginner Implicit Differentiation

Differentiate all parts of the expressions below. Treat y as a function of x.

Remember, (y) =    &   (x) = 1

a) y + 2x + 1

So we differentiate each part with respect to x. We can treat it like we are "multiplying out the brackets."

(y + 2x + 1)

(y) +(2x) + (1)

+ 2 + 0

Note: With Implicit Differentiation, we leave our answers with in them

b) 2y + 3x2

Again apply to all terms in the equation.

(2y + 3x2)

= (2y) + (3x2)

Note: Do not make the mistake of differentiating (2y) = 2 - that is incorrect.

See the "y" as a function of xy(x) and that may remind you how to differentiate it.

= 2(y) + 6x

= 2 + 6x

c)y2

Be careful. Remember y is a function of x. i.e. y(x). We cannot do (y2) = 2y

Apply ()

Apply (y2)

Here we have the derivative of a function to a power. We will use the Power Rule for Functions.

#### Power Rule For Functions

(f(x)n) = n.f(x)n-1f(x)

Here we have:

(yn) - you can think of y as y(x)

= 2.y2-1.y

= 2y.

### Implicit Derivative

Having used implicit differentiation, we are often left with a in amongst other terms, for example, an answer like,

4 = 2x + 2(),

We can use basic algebra to rearrange these types of equations as if we were solving for , trying to get the on one side of the equation on its own,

4 - 2x = 2()

4 - 2x
2
=

= 2 - x