Z%3f(x, y)%3D 5(x^3)(y^2)-7x^2 +8y-4. Find the value of the second and mixed second partial derivatives at (2, 3); that is find frx(2,3), fyy(2,3), fxy(2,3). _fxx D fyy = fxy%= %3D %3D %3D
Z%3f(x, y)%3D 5(x^3)(y^2)-7x^2 +8y-4. Find the value of the second and mixed second partial derivatives at (2, 3); that is find frx(2,3), fyy(2,3), fxy(2,3). _fxx D fyy = fxy%= %3D %3D %3D
Calculus: Early Transcendentals
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Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Second Partial Derivatives of the Function
Given the function:
\[ Z = f(x, y) = 5(x^3)(y^2) - 7x^2 + 8y - 4 \]
We are to find the value of the second and mixed second partial derivatives at the point (2, 3). Specifically, we need to determine \( f_{xx}(2,3) \), \( f_{yy}(2,3) \), and \( f_{xy}(2,3) \).
### Step-by-Step Solution:
#### 1. Compute \( f_{xx} \):
To find \( f_{xx} \), we first need the first partial derivative of \( f \) with respect to \( x \), \( f_x \):
\[ f_x = \frac{\partial}{\partial x} [5(x^3)(y^2) - 7x^2 + 8y - 4] \]
\[ f_x = 5(y^2) \frac{\partial}{\partial x} (x^3) - 7 \frac{\partial}{\partial x} (x^2) + 8 \frac{\partial}{\partial x} (y) - 4 \frac{\partial}{\partial x} (1) \]
\[ f_x = 5(y^2)(3x^2) - 7(2x) + 0 - 0 \]
\[ f_x = 15x^2 y^2 - 14x \]
Now, take the second partial derivative \( f_{xx} \):
\[ f_{xx} = \frac{\partial}{\partial x} (15x^2 y^2 - 14x) \]
\[ f_{xx} = 15(y^2) \frac{\partial}{\partial x} (x^2) - 14 \frac{\partial}{\partial x} (x) \]
\[ f_{xx} = 15(y^2)(2x) - 14 \]
\[ f_{xx} = 30xy^2 - 14 \]
Evaluate at the point (2, 3):
\[ f_{xx}(2,3) = 30(2)(3^2) - 14 \]
\[ f_{xx}(2,3) = 30(2)(9) - 14 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdd93d516-b23d-479f-a2e8-0184d486bfe0%2Faad1f3e4-d8ac-46c5-b38a-3777e0f09831%2Fgftoqsp.jpeg&w=3840&q=75)
Transcribed Image Text:### Second Partial Derivatives of the Function
Given the function:
\[ Z = f(x, y) = 5(x^3)(y^2) - 7x^2 + 8y - 4 \]
We are to find the value of the second and mixed second partial derivatives at the point (2, 3). Specifically, we need to determine \( f_{xx}(2,3) \), \( f_{yy}(2,3) \), and \( f_{xy}(2,3) \).
### Step-by-Step Solution:
#### 1. Compute \( f_{xx} \):
To find \( f_{xx} \), we first need the first partial derivative of \( f \) with respect to \( x \), \( f_x \):
\[ f_x = \frac{\partial}{\partial x} [5(x^3)(y^2) - 7x^2 + 8y - 4] \]
\[ f_x = 5(y^2) \frac{\partial}{\partial x} (x^3) - 7 \frac{\partial}{\partial x} (x^2) + 8 \frac{\partial}{\partial x} (y) - 4 \frac{\partial}{\partial x} (1) \]
\[ f_x = 5(y^2)(3x^2) - 7(2x) + 0 - 0 \]
\[ f_x = 15x^2 y^2 - 14x \]
Now, take the second partial derivative \( f_{xx} \):
\[ f_{xx} = \frac{\partial}{\partial x} (15x^2 y^2 - 14x) \]
\[ f_{xx} = 15(y^2) \frac{\partial}{\partial x} (x^2) - 14 \frac{\partial}{\partial x} (x) \]
\[ f_{xx} = 15(y^2)(2x) - 14 \]
\[ f_{xx} = 30xy^2 - 14 \]
Evaluate at the point (2, 3):
\[ f_{xx}(2,3) = 30(2)(3^2) - 14 \]
\[ f_{xx}(2,3) = 30(2)(9) - 14 \
Expert Solution
Step 1
We differentiate with respect to x twice, y will be treated as a constant.
Step 2
We differentiate with respect to y twice, x will be treated as a constant.
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