Given z = 2xy^4 + 7x^3y, the values of the second-order direct partial derivatives are, Given z = 3x^2y^3, when x=4, y=1, the values of the second-order cross partial derivatives a... Given z = 2xy^4 + 7x^3y, the values of the second-order direct partial derivatives are, Given z = 3x^2y^3, when x=4, y=1, the values of the second-order cross partial derivatives are, The Taylor series for tan x is given as, The Maclaurin series for f(x) = e^x is, This Taylor Series expansion x - (x^2/2) + (x^3/3) - (x^4/4) at x=0 is of,
Understand the Problem
The question involves solving problems related to partial derivatives, Taylor series, and Maclaurin series expansion in calculus. It requires knowledge of second-order derivatives and series expansions.
Answer
The values of the second-order direct partial derivatives are \( z_{xx} = 42xy \), \( z_{yy} = 24y^2 \), and for cross derivatives, \( z_{xy} = 12 \), \( z_{yx} = 9 \).
Answer for screen readers
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The values of the second-order direct partial derivatives are ( z_{xx} = 42xy ), ( z_{yy} = 24y^2 ).
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For the second-order cross partial derivatives when ( x = 4, y = 1 ):
- ( z_{xy} = 12 )
- ( z_{yx} = 9 ).
Steps to Solve
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Identify the Variables and Functions for Direct Partial Derivatives
Given the function ( z = 2xy^4 + 7x^3y ), we need to find the second-order direct partial derivatives: ( z_{xx} ) and ( z_{yy} ).
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Calculate the First-Order Partial Derivatives
First, we find the first-order derivatives.
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The partial derivative with respect to ( x ): $$ z_x = \frac{\partial}{\partial x}(2xy^4 + 7x^3y) = 2y^4 + 21x^2y $$
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The partial derivative with respect to ( y ): $$ z_y = \frac{\partial}{\partial y}(2xy^4 + 7x^3y) = 8xy^3 + 7x^3 $$
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Calculate the Second-Order Direct Partial Derivatives
Now we compute the second derivatives:
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For ( z_{xx} ): $$ z_{xx} = \frac{\partial}{\partial x}(z_x) = \frac{\partial}{\partial x}(2y^4 + 21x^2y) = 42xy $$
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For ( z_{yy} ): $$ z_{yy} = \frac{\partial}{\partial y}(z_y) = \frac{\partial}{\partial y}(8xy^3 + 7x^3) = 24xy^2 $$
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Collecting the Results of Direct Partial Derivatives
Thus, the values of the second-order direct partial derivatives are:
- ( z_{xx} = 42xy )
- ( z_{yy} = 24xy^2 )
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Given Values for Second-Order Cross Partial Derivatives
For the function ( z = 3xy^3 ) at ( x=4 ) and ( y=1 ):
- Calculate ( z_{xy} ) and ( z_{yx} ): $$ z_{xy} = \frac{\partial}{\partial y}(z_x) = 3x^3 $$ $$ z_{yx} = \frac{\partial}{\partial x}(z_y) = 9y^2 $$
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Evaluate at ( x=4, y=1 )
Finally, substituting ( x=4 ) and ( y=1 ):
- ( z_{xy} = 3(4)(1)^3 = 12 )
- ( z_{yx} = 9(1)^2 = 9 )
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The values of the second-order direct partial derivatives are ( z_{xx} = 42xy ), ( z_{yy} = 24y^2 ).
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For the second-order cross partial derivatives when ( x = 4, y = 1 ):
- ( z_{xy} = 12 )
- ( z_{yx} = 9 ).
More Information
The second-order direct partial derivatives give insight into how the function behaves with respect to changes in ( x ) and ( y ) independently, while the cross partial derivatives demonstrate how changes in one variable affect the changes in another variable.
Tips
- Forgetting to differentiate correctly: Ensure to apply the product rule correctly when variables are multiplied.
- Neglecting to evaluate: Always substitute the values back into derivatives to avoid errors in the final outcome.
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