Solved on Jan 31, 2024

Find the gradient of the function f(x,y)=xexy2+cosy2f(x, y) = x e^{xy^2} + \cos y^2.

STEP 1

Assumptions
1. The function given is f(x,y)=xexy2+cos(y2)f(x, y) = x e^{x y^2} + \cos(y^2).
2. The gradient of a function, denoted as f\nabla f, is a vector that consists of the partial derivatives of the function with respect to each variable.
3. The gradient of f(x,y)f(x, y) is a two-dimensional vector since ff is a function of two variables, xx and yy.
4. The gradient vector is given by f(x,y)=(fx,fy)\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right).

STEP 2

First, we will find the partial derivative of the function f(x,y)f(x, y) with respect to xx.
fx=x(xexy2)+x(cos(y2))\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( x e^{x y^2} \right) + \frac{\partial}{\partial x} \left( \cos(y^2) \right)

STEP 3

Calculate the partial derivative of the first term with respect to xx using the product rule, which states that x(uv)=uvx+vux\frac{\partial}{\partial x} (u v) = u \frac{\partial v}{\partial x} + v \frac{\partial u}{\partial x}, where uu and vv are functions of xx.
Let u=xu = x and v=exy2v = e^{x y^2}, then we have:
x(xexy2)=xx(exy2)+exy2x(x)\frac{\partial}{\partial x} \left( x e^{x y^2} \right) = x \frac{\partial}{\partial x} \left( e^{x y^2} \right) + e^{x y^2} \frac{\partial}{\partial x} (x)

STEP 4

Calculate the derivative of exy2e^{x y^2} with respect to xx.
x(exy2)=y2exy2\frac{\partial}{\partial x} \left( e^{x y^2} \right) = y^2 e^{x y^2}

STEP 5

Calculate the derivative of xx with respect to xx.
x(x)=1\frac{\partial}{\partial x} (x) = 1

STEP 6

Substitute the derivatives from steps 4 and 5 into the product rule from step 3.
x(xexy2)=xy2exy2+exy2(1)\frac{\partial}{\partial x} \left( x e^{x y^2} \right) = x y^2 e^{x y^2} + e^{x y^2} (1)

STEP 7

Simplify the expression.
x(xexy2)=xy2exy2+exy2\frac{\partial}{\partial x} \left( x e^{x y^2} \right) = x y^2 e^{x y^2} + e^{x y^2}

STEP 8

Calculate the partial derivative of the second term with respect to xx. Since cos(y2)\cos(y^2) does not depend on xx, its derivative with respect to xx is zero.
x(cos(y2))=0\frac{\partial}{\partial x} \left( \cos(y^2) \right) = 0

STEP 9

Combine the derivatives from steps 7 and 8 to find the partial derivative of f(x,y)f(x, y) with respect to xx.
fx=xy2exy2+exy2+0\frac{\partial f}{\partial x} = x y^2 e^{x y^2} + e^{x y^2} + 0

STEP 10

Simplify the expression by factoring out exy2e^{x y^2}.
fx=exy2(xy2+1)\frac{\partial f}{\partial x} = e^{x y^2} (x y^2 + 1)

STEP 11

Next, we will find the partial derivative of the function f(x,y)f(x, y) with respect to yy.
fy=y(xexy2)+y(cos(y2))\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( x e^{x y^2} \right) + \frac{\partial}{\partial y} \left( \cos(y^2) \right)

STEP 12

Calculate the partial derivative of the first term with respect to yy using the chain rule, which states that y(uv)=uvy+vuy\frac{\partial}{\partial y} (u v) = u \frac{\partial v}{\partial y} + v \frac{\partial u}{\partial y}, where uu and vv are functions of yy.
Let u=xu = x and v=exy2v = e^{x y^2}, then we have:
y(xexy2)=xy(exy2)+exy2y(x)\frac{\partial}{\partial y} \left( x e^{x y^2} \right) = x \frac{\partial}{\partial y} \left( e^{x y^2} \right) + e^{x y^2} \frac{\partial}{\partial y} (x)

STEP 13

Since xx is a constant with respect to yy, its derivative with respect to yy is zero.
y(x)=0\frac{\partial}{\partial y} (x) = 0

STEP 14

Calculate the derivative of exy2e^{x y^2} with respect to yy using the chain rule.
y(exy2)=y(xy2)exy2\frac{\partial}{\partial y} \left( e^{x y^2} \right) = \frac{\partial}{\partial y} \left( x y^2 \right) e^{x y^2}

STEP 15

Calculate the derivative of xy2x y^2 with respect to yy.
y(xy2)=2xy\frac{\partial}{\partial y} \left( x y^2 \right) = 2xy

STEP 16

Substitute the derivative from step 15 into the chain rule from step 14.
y(exy2)=2xyexy2\frac{\partial}{\partial y} \left( e^{x y^2} \right) = 2xy e^{x y^2}

STEP 17

Substitute the derivatives from steps 13 and 16 into the chain rule from step 12.
y(xexy2)=x(2xyexy2)+exy2(0)\frac{\partial}{\partial y} \left( x e^{x y^2} \right) = x (2xy e^{x y^2}) + e^{x y^2} (0)

STEP 18

Simplify the expression.
y(xexy2)=2x2yexy2\frac{\partial}{\partial y} \left( x e^{x y^2} \right) = 2x^2y e^{x y^2}

STEP 19

Calculate the partial derivative of the second term with respect to yy using the chain rule.
y(cos(y2))=sin(y2)y(y2)\frac{\partial}{\partial y} \left( \cos(y^2) \right) = -\sin(y^2) \frac{\partial}{\partial y} (y^2)

STEP 20

Calculate the derivative of y2y^2 with respect to yy.
y(y2)=2y\frac{\partial}{\partial y} (y^2) = 2y

STEP 21

Substitute the derivative from step 20 into the chain rule from step 19.
y(cos(y2))=sin(y2)(2y)\frac{\partial}{\partial y} \left( \cos(y^2) \right) = -\sin(y^2) (2y)

STEP 22

Simplify the expression.
y(cos(y2))=2ysin(y2)\frac{\partial}{\partial y} \left( \cos(y^2) \right) = -2y \sin(y^2)

STEP 23

Combine the derivatives from steps 18 and 22 to find the partial derivative of f(x,y)f(x, y) with respect to yy.
fy=2x2yexy22ysin(y2)\frac{\partial f}{\partial y} = 2x^2y e^{x y^2} - 2y \sin(y^2)

STEP 24

Now that we have both partial derivatives, we can write the gradient of the function f(x,y)f(x, y).
f(x,y)=(fx,fy)\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)

STEP 25

Substitute the partial derivatives from steps 10 and 23 into the gradient vector.
f(x,y)=(exy2(xy2+1),2x2yexy22ysin(y2))\nabla f(x, y) = \left( e^{x y^2} (x y^2 + 1), 2x^2y e^{x y^2} - 2y \sin(y^2) \right)
The gradient of the function f(x,y)=xexy2+cos(y2)f(x, y) = x e^{x y^2} + \cos(y^2) is:
f(x,y)=(exy2(xy2+1),2x2yexy22ysin(y2))\nabla f(x, y) = \left( e^{x y^2} (x y^2 + 1), 2x^2y e^{x y^2} - 2y \sin(y^2) \right)

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