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#BEEBETTER at www.tutorbee.tv Differentiation Implicit Differentiation Level 8 12 IMPLICIT DIFFERENTIATION The equation of an explicit function has its dependent variable on one side of the equal sign and the independent variable(s) on the other side: = $ + 2 − 1 In the equation of an implicit function dependent variable is not alone: 2 $ = 2 Recall that the derivative of the is or /. Particularly in the last notation, we refer to the derivative as “the derivative of with respect to .” When taking the derivative of an implicit function, whenever you take the derivative of , you must multiply by / or (by Chain Rule). So, the derivative of $ with respect to is: $ . = 2 ∙ = 2 But the derivative to $ with respect to is: ( $ )′ = 2 ∙ After taking the derivative of both sides, isolate for / to get the derivative. This process is called implicit differentiation. EXAMPLE: Find the derivative of the circle: $ + $ = 25. SOLUTION: We’ll take the derivatives of both sides and isolate for /: 2 + 2 ∙ =0 2 ∙ = −2 / = −/ EXAMPLE: Find the derivative of + = 2 − 1. SOLUTION: Notice that the first term is a product of two functions: and . We’ll need to use product rule to differentiate: 1 + ∙ ′ + ′ = 2 + 0

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#BEEBETTER at www.tutorbee.tv

Differentiation Implicit Differentiation

Level 8

12

IMPLICIT DIFFERENTIATION The equation of an explicit function has its dependent variable on one side of the equal sign and the independent variable(s) on the other side:

𝑦 = 𝑥$ + 2𝑥 − 1

In the equation of an implicit function dependent variable is not alone:

2𝑥$𝑦 −𝑥𝑦= 2𝑥

Recall that the derivative of the 𝑦 is 𝑦′ or 𝑑𝑦/𝑑𝑥. Particularly in the last notation, we refer to the derivative as “the derivative of 𝑦 with respect to 𝑥.” When taking the derivative of an implicit function, whenever you take the derivative of 𝑦, you must multiply by 𝑑𝑦/𝑑𝑥 or 𝑦′ (by Chain Rule). So, the derivative of 𝑥$ with respect to 𝑥 is:

𝑥$ . = 2𝑥 ∙𝑑𝑥𝑑𝑥

= 2𝑥

But the derivative to 𝑦$ with respect to 𝑥 is:

(𝑦$)′ = 2𝑦 ∙𝑑𝑦𝑑𝑥

After taking the derivative of both sides, isolate for 𝑑𝑦/𝑑𝑥 to get the derivative. This process is called implicit differentiation. EXAMPLE: Find the derivative of the circle: 𝑥$ + 𝑦$ = 25. SOLUTION: We’ll take the derivatives of both sides and isolate for 𝑑𝑦/𝑑𝑥:

2𝑥 + 2𝑦 ∙𝑑𝑦𝑑𝑥

= 0

2𝑦 ∙𝑑𝑦𝑑𝑥

= −2𝑥

𝑑𝑦/𝑑𝑥 = −𝑥/𝑦

EXAMPLE: Find the derivative of 𝑥𝑦 + 𝑦 = 2𝑥 − 1. SOLUTION: Notice that the first term is a product of two functions: 𝑥 and 𝑦. We’ll need to use product rule to differentiate:

1 𝑦 + 𝑥 ∙ 𝑦′ + 𝑦′ = 2 + 0

#BEEBETTER at www.tutorbee.tv

Differentiation Implicit Differentiation

Level 8

13

Move 𝑦 to the right side and factor out the 𝑦′:

𝑦′ 𝑥 + 1 = 2 − 𝑦

Now isolate for 𝑦′:

𝑦′ =− 𝑦 − 2𝑥 + 1

EXAMPLE: Find the derivative of 𝑥𝑦 = 4 both implicity and explicitly. SOLUTION: We can solve for 𝑦 to obtain an explicit equation:

𝑦 =4𝑥= 4𝑥56

Then take the derivative:

𝑦. = −4𝑥5$ = −4𝑥$

Implicitly, we could use product rule:

1 𝑦 + 𝑦. 𝑥 = 0𝑥𝑦. = −𝑦𝑦. = −𝑦/𝑥

This may look different but recall that 𝑦 = 4/𝑥. We can subsitute that explicit equation into the derivative and see that it is the same expression:

𝑦. =−4𝑥𝑥

= −4𝑥$

Manipulating Functions You can manipulate functions before taking the derivative in order to avoid quotient rule and get rid of square roots. For example:

𝑦 − 1𝑥

=𝑥𝑦3

Square both sides to obtain:

𝑦$ − 2𝑦 + 1𝑥$

=𝑥𝑦9

Then cross multiply:

9𝑦$ − 18𝑦 + 9 = 𝑥:𝑦

This is an easier function to differentiate.