day 10 - implicit differentiation - 10 - implicit آ  differentiation implicit differentiation

Download Day 10 - Implicit Differentiation - 10 - Implicit  آ  Differentiation Implicit Differentiation

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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 − 𝑥 𝑦 = 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 ∙ 𝑑𝑥 𝑑𝑥

    = 2𝑥

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

    (𝑦$)′ = 2𝑦 ∙ 𝑑𝑦 𝑑𝑥

    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 ∙ 𝑑𝑦 𝑑𝑥

    = 0

    2𝑦 ∙ 𝑑𝑦 𝑑𝑥

    = −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 − 2 𝑥 + 1

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

    𝑦 = 4 𝑥 = 4𝑥56

    Then take the derivative:

    𝑦. = −4𝑥5$ = − 4 𝑥$

    Implicitly, we could use product rule:

    1 𝑦 + 𝑦. 𝑥 = 0 𝑥𝑦. = −𝑦 𝑦. = −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. = −4 𝑥 𝑥

    = − 4 𝑥$

    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 − 1 𝑥

    = 𝑥𝑦 3

    Square both sides to obtain:

    𝑦$ − 2𝑦 + 1 𝑥$

    = 𝑥𝑦 9

    Then cross multiply:

    9𝑦$ − 18𝑦 + 9 = 𝑥:𝑦

    This is an easier function to differentiate.