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 y^{2} - x ^{2} = 3x + 1, we can write it explicitly like, y = **√**(x^{2} + 3x + 1).

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

x - cos(x) = y^{4} + 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*.

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 = 3x^{2} + 2x + 1

(y) = (3x^{2}) + (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.

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)

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 + 3x^{2}

Again apply to all terms in the equation.

(2y + 3x^{2})

= (2y) + (3x^{2})

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

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

= 2(y) + 6x

= 2 + 6x

c)y^{2}

Be careful. Remember *y* is a function of *x*. i.e. *y(x)*. We cannot do (*y*^{2}) = 2*y*

Apply ()

Apply (y^{2})

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

(f(x)^{n}) = n.f(x)^{n-1}f(x)

Here we have:

(y^{n}) - you can think of *y* as *y(x)*

= 2.y^{2-1}.y

= 2y.

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