## 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}}} .

## Video on Implicit Differentiation

## 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