Find the derivative of all orders of the function y = x^4/4 + 4/9 x^3 - x^2 + 3x - 8.

Understand the Problem

The question is asking to find the derivatives of a given function up to the fifth order. This involves applying differentiation rules to calculate each derivative step by step.

Answer

- $ y' = x^3 + \frac{4}{3}x^2 - 2x + 3 $ - $ y'' = 3x^2 + \frac{8}{3}x - 2 $ - $ y''' = 6x + \frac{8}{3} $ - $ y^{(4)} = 6 $ - $ y^{(5)} = 0 $

Steps to Solve

Find the first derivative

To differentiate ( y = \frac{x^4}{4} + \frac{4}{9}x^3 - x^2 + 3x - 8 ), apply the power rule for each term.

Using the power rule ( \frac{d}{dx}x^n = nx^{n-1} ):

$$ y' = \frac{1}{4} \cdot 4x^{4-1} + \frac{4}{9} \cdot 3x^{3-1} - 2x^{2-1} + 3 $$ $$ y' = x^3 + \frac{4}{3}x^2 - 2x + 3 $$

Find the second derivative

Now differentiate ( y' ):

$$ y'' = 3x^{2} + \frac{4}{3} \cdot 2x^{1} - 2 $$ $$ y'' = 3x^2 + \frac{8}{3}x - 2 $$

Find the third derivative

Differentiate ( y'' ):

$$ y''' = 6x + \frac{8}{3} $$

Find the fourth derivative

Differentiate ( y''' ):

$$ y^{(4)} = 6 $$

Find the fifth derivative

Differentiate ( y^{(4)} ):

$$ y^{(5)} = 0 $$

The derivatives are as follows:

( y' = x^3 + \frac{4}{3}x^2 - 2x + 3 )
( y'' = 3x^2 + \frac{8}{3}x - 2 )
( y''' = 6x + \frac{8}{3} )
( y^{(4)} = 6 )
( y^{(5)} = 0 )

More Information

The function provided is a polynomial of degree 4. Its derivatives reduce the degree by one with each differentiation until reaching a constant (fourth derivative) and then zero (fifth derivative).

Tips

Forgetting to apply the power rule correctly, especially with coefficients in front of the variables.
Confusing the signs when differentiating; ensure you watch for negative coefficients.
Neglecting constant terms or missing differentiation of constants (any constant has a derivative of zero).

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