Unit 2 - Objective 4 - Higher Derivatives



The derivative of the function is the 1st derivative.

The derivative of the 1st derivative is the 2nd derivative.

The derivative of the 2nd derivative is the 3rd derivative and so on.




Problem 1:
Find the 2nd derivative of
The first derivative is
The second derivative is

Problem 2:

Evaluate the 2nd derivative of the function for the given value of x

   y = 3x (x² + 2)³ at the point x = 1

You need to find the 1st derivative before you can find the 2nd derivative

    = 3x (x² + 2)³ + (x² + 2)³ (3x)

    = 3x (3) (x² + 2)² (2x) + (x² + 2)³ (3)

    = 18x² (x² + 2)² + 3 (x² + 2)³

You can factor out (x² + 2)³

    = (x² + 2)² [18x² + 3 (x² + 2)]

    = (x² + 2)² [18x² + 3x² +6]

    = (x² + 2)² [21x² + 6]


Now you can find the 2nd derivative

    = (x² + 2)² [x² +6] + [21x² + 6] (x² + 2)²

    = (x² + 2)² (42x) + (21x² + 6) (2) (x² + 2) (2x)

    = 42x (x² + 2)² +4x (21x² + 6) (x² + 2)


Evaluate the 2nd derivative at x = 1

    = 42 (1) (1² + 2)² + 4 (1) (21 (1)² +6) (1² + 2)

    = 42 (9) + 4 (27) (3)

    = 378 + 324

    = 702




Implicit differentiation can also produce derivatives of a higher order.



Find the 2nd derivative if 2x³ - 3y² = 7

(2x³) - (3y²) = (7)

6x² - 6y = 0

x² - y = 0

=

We now apply the quotient rule to find the 2nd derivative.

We now substitute the expression for the 1st derivative to express the 2nd derivative in terms of x and y.


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