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Find the gradient of the function. Assume the variables are restricted to a domain on which the function is defined. z=2xe2yx+2y. Enter an answer in each field. Enclose numerators and denominators in parentheses. For example, (a−b)/(1+n).

Respuesta :

mavila18

Answer:

[tex]grad(z(x,y))=2[e^{2xy}+2xye^{2xy}]\hat{i}+[4x^{2}e^{2xy}+2]\hat{j}[/tex]

Step-by-step explanation:

[tex]z=2xe^{2yx}+2y[/tex]

The gradient of a function is

[tex]grad(z(x,y))=D_{x}z(x,y)\hat{i}+D_{y}z(x,y)\hat{j}[/tex]

where Dx means derivative over x and the same for Dy. Hence we have

[tex]grad(z(x,y))=2[e^{2xy}+2xye^{2xy}]\hat{i}+[4x^{2}e^{2xy}+2]\hat{j}[/tex]

where we have used derivative of a product anf of an exponential function

I hope this is useful for you

regards

Space

Answer:

[tex]\displaystyle \nabla z = 2e^\big{2yx} \Big( 2xy + 1 \Big) \hat{\i} + 2 \Big( 2x^2e^\big{2yx} + 1 \Big) \hat{\j}[/tex]

General Formulas and Concepts:
Calculus

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Multivariable Calculus

Differentiation

  • Partial Derivatives
  • Derivative Notation

Partial Derivative Rule [Chain Rule]:                                                                    [tex]\displaystyle \frac{\partial f}{\partial t} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t} + \frac{\partial f}{\partial z} \frac{\partial z}{\partial t}[/tex]

Gradient:                                                                                                               [tex]\displaystyle \nabla f(x, y, z) = \frac{\partial f}{\partial x} \hat{\i} + \frac{\partial f}{\partial y} \hat{\j} + \frac{\partial f}{\partial z} \hat{\text{k}}[/tex]

Gradient Property [Addition/Subtraction]:                                                          [tex]\displaystyle \nabla \big[ f(x) + g(x) \big] = \nabla f(x) + \nabla g(x)[/tex]

Gradient Property [Multiplied Constant]:                                                            [tex]\displaystyle \nabla \big[ \alpha f(x) \big] = \alpha \nabla f(x)[/tex]

Gradient Rule [Product Rule]:                                                                              [tex]\displaystyle \nabla \big[ f(x)g(x) \big] = f(x) \nabla g(x) + \nabla f(x) g(x)[/tex]

Step-by-step explanation:

Step 1: Define

Identify.

[tex]\displaystyle z = 2xe^\big{2yx} + 2y[/tex]

Step 2: Find Gradient

  1. [Function] Differentiate [Gradient]:                                                            [tex]\displaystyle \nabla z = \frac{\partial}{\partial x} \bigg[ 2xe^\big{2yx} + 2y \bigg] \hat{\i} + \frac{\partial}{\partial y} \bigg[ 2xe^\big{2yx} + 2y \bigg] \hat{\j}[/tex]
  2. [Gradient] Rewrite [Gradient Property - Multiplied Constant]:                  [tex]\displaystyle \nabla z = 2 \frac{\partial}{\partial x} \bigg[ xe^\big{2yx} + y \bigg] \hat{\i} + 2 \frac{\partial}{\partial y} \bigg[ xe^\big{2yx} + y \bigg] \hat{\j}[/tex]
  3. [Gradient] Rewrite [Gradient Property - Addition/Subtraction]:                [tex]\displaystyle \nabla z = 2 \bigg[ \frac{\partial}{\partial x} \bigg( xe^\big{2yx} \bigg) + \frac{\partial}{\partial x}(y) \bigg] \hat{\i} + 2 \bigg[ \frac{\partial}{\partial y} \bigg( xe^\big{2yx} \bigg) + \frac{\partial}{\partial y}(y) \bigg] \hat{\j}[/tex]
  4. [Gradient] Differentiate [Derivative Rule - Basic Power Rule]:                  [tex]\displaystyle \nabla z = 2 \frac{\partial}{\partial x} \bigg( xe^\big{2yx} \bigg) \hat{\i} + 2 \bigg[ \frac{\partial}{\partial y} \bigg( xe^\big{2yx} \bigg) + 1 \bigg] \hat{\j}[/tex]
  5. [Gradient] Rewrite [Gradient Rule - Product Rule]:                                    [tex]\displaystyle \nabla z = 2 \bigg[ \frac{\partial}{\partial x} (x) e^\big{2yx} + x \frac{\partial}{\partial x} \bigg( e^\big{2yx} \bigg) \bigg] \hat{\i} + 2 \bigg[ \frac{\partial}{\partial y} (x) e^\big{2yx} + x \frac{\partial}{\partial y} \bigg( e^\big{2yx} \bigg) + 1 \bigg] \hat{\j}[/tex]
  6. [Gradient] Differentiate [Derivative Rule - Basic Power Rule]:                  [tex]\displaystyle \nabla z = 2 \bigg[ e^\big{2yx} + x \frac{\partial}{\partial x} \bigg( e^\big{2yx} \bigg) \bigg] \hat{\i} + 2 \bigg[ x \frac{\partial}{\partial y} \bigg( e^\big{2yx} \bigg) + 1 \bigg] \hat{\j}[/tex]
  7. [Gradient] Differentiate [Partial Derivative Rule - Chain Rule]:                  [tex]\displaystyle \nabla z = 2 \bigg[ e^\big{2yx} + xe^\big{2yx} \frac{\partial}{\partial x} (2yx) \bigg] \hat{\i} + 2 \bigg[ xe^\big{2yx} \frac{\partial}{\partial y}(2yx) + 1 \bigg] \hat{\j}[/tex]
  8. [Gradient] Rewrite [Gradient Property - Multiplied Constant]:                  [tex]\displaystyle \nabla z = 2 \bigg[ e^\big{2yx} + 2xe^\big{2yx} \frac{\partial}{\partial x} (yx) \bigg] \hat{\i} + 2 \bigg[ 2xe^\big{2yx} \frac{\partial}{\partial y}(yx) + 1 \bigg] \hat{\j}[/tex]
  9. [Gradient] Differentiate [Derivative Rule - Basic Power Rule]:                  [tex]\displaystyle \nabla z = 2 \bigg[ e^\big{2yx} + 2xye^\big{2yx} \bigg] \hat{\i} + 2 \bigg[ 2x^2e^\big{2yx} + 1 \bigg] \hat{\j}[/tex]
  10. [Gradient] Simplify:                                                                                       [tex]\displaystyle \nabla z = 2e^\big{2yx} \Big( 1 + 2xy \Big) \hat{\i} + 2 \Big( 2x^2e^\big{2yx} + 1 \Big) \hat{\j}[/tex]
  11. [Gradient] Rewrite:                                                                                       [tex]\displaystyle \nabla z = 2e^\big{2yx} \Big( 2xy + 1 \Big) \hat{\i} + 2 \Big( 2x^2e^\big{2yx} + 1 \Big) \hat{\j}[/tex]

∴ we have found the gradient of the given function z = f(x, y).

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Learn more about gradient: https://brainly.com/question/7369278

Learn more about multivariable calculus: https://brainly.com/question/17433118

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Topic: Multivariable Calculus

Unit: Directional Derivatives

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