# Implicit Differentiation

## Introduction to Implicit Differentiation

Implicit differentiation is necessary when we take the derivative of two variables not only just one.

An example of implicit differentiation is as follows: x 2 + y 2 = 2 {\displaystyle x^{2}+y^{2}=2}

## Working Through an Implicit Differentiation Question

1. Identify that it is an implicit differentiation question. (There will be two variables. The "y" variable must be to a greater power than 1 or otherwise locked up in a non-linear function, such as ln(y).)

2. Differentiate both sides of the equals sign with d/dx. With the "y" variable, make sure to differentiate and multiply that new value by dy/dx -- you are applying the chain rule -- (i.e. the dy/dx remains "glued" to the y variable and its coefficient and power even after it has been differentiated)

2a. For x values, when differentiating with d/dx, just use the differentiation rules found on Simple Differentiation

3. Solve for dy/dx using any algebraic and arithmetic methods.

## Note on Implicit Differentiation

There will be times, especially in word problems, where you will be required to take the derivative of an equation with respect to something other than x. You will be able to notice this, if there the question asks how fast something is moving, or more simply, the question will tell you with which variable o take the derivative with.

When taking the derivative of a variable, with respect to a different variable you must always include the d(variable)/d(variable) as a coefficient.

For example: Take the derivative of x 2 + y 2 = 1 {\displaystyle x^{2}+y^{2}=1} with respect to time (t):

x 2 + y 2 = 1 {\displaystyle x^{2}+y^{2}=1}

and so,

d ( x 2 + y 2 ) d t = d ( 1 ) d t {\displaystyle {\frac {d(x^{2}+y^{2})}{dt}}={\frac {d(1)}{dt}}} ,

which means

2 x d x d t + 2 y d y d t = 0 {\displaystyle 2x{\frac {dx}{dt}}+2y{\frac {dy}{dt}}=0} ,

giving

d y d t = − x y d x d t {\displaystyle {\frac {dy}{dt}}=-{\frac {x}{y}}{\frac {dx}{dt}}} .

## Worked Out Problem

1.

x 2 + y 2 = 2 {\displaystyle x^{2}+y^{2}=2}

d / d x ( x 2 + y 2 ) = d / d x ( 2 ) {\displaystyle d/dx(x^{2}+y^{2})=d/dx(2)}

2 x + 2 y d y / d x = 0 {\displaystyle 2x+2ydy/dx=0}

2 y d y / d x = − 2 x {\displaystyle 2ydy/dx=-2x}

d y / d x = − 2 x / 2 y {\displaystyle dy/dx=-2x/2y}

2.

x 2 + y 3 = x 3 + y 2 + 5 {\displaystyle x^{2}+y^{3}=x^{3}+y^{2}+5}

d / d x ( x 2 + y 3 ) = d / d x ( x 2 + y 2 + 5 ) {\displaystyle d/dx(x^{2}+y^{3})=d/dx(x^{2}+y^{2}+5)}

2 x + 3 y 2 d y / d x = 3 x 2 + 2 y d y / d x + 0 {\displaystyle 2x+3y^{2}dy/dx=3x^{2}+2ydy/dx+0}

2 x − 3 x 2 = 2 y d y / d x − 3 y 2 d y / d x {\displaystyle 2x-3x^{2}=2ydy/dx-3y^{2}dy/dx}

d y / d x = ( 2 x − 3 x 2 ) / ( 2 y − 3 y 2 ) {\displaystyle dy/dx=(2x-3x^{2})/(2y-3y^{2})}

## Practical Uses for Implicit Differentiation

Implicit differentiation (for this course) can be found as a simple equation, but it will more use when figuring our related rates problems or optimization problems.

Here is an example of one of these situations:

## Practice Problems

To find many practice problems please see Practice Problems