Solved on Jan 31, 2024

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

STEP 1

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

STEP 2

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

STEP 3

Calculate the partial derivative of the first term with respect to $x$ using the product rule, which states that $\frac{\partial}{\partial x} (u v) = u \frac{\partial v}{\partial x} + v \frac{\partial u}{\partial x}$ , where $u$ and $v$ are functions of $x$ .
Let $u = x$ and $v = e^{x y^2}$ , then we have:
$\frac{\partial}{\partial x} \left( x e^{x y^2} \right) = x \frac{\partial}{\partial x} \left( e^{x y^2} \right) + e^{x y^2} \frac{\partial}{\partial x} (x)$

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

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

STEP 10

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

STEP 11

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

STEP 12

Calculate the partial derivative of the first term with respect to $y$ using the chain rule, which states that $\frac{\partial}{\partial y} (u v) = u \frac{\partial v}{\partial y} + v \frac{\partial u}{\partial y}$ , where $u$ and $v$ are functions of $y$ .
Let $u = x$ and $v = e^{x y^2}$ , then we have:
$\frac{\partial}{\partial y} \left( x e^{x y^2} \right) = x \frac{\partial}{\partial y} \left( e^{x y^2} \right) + e^{x y^2} \frac{\partial}{\partial y} (x)$

STEP 13

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

STEP 14

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

STEP 15

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

STEP 16

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

STEP 17

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

STEP 18

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

STEP 19

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

STEP 20

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

STEP 21

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

STEP 22

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

STEP 23

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

STEP 24

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

STEP 25

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

Was this helpful?

Find the gradient of the function f(x,y)=xexy2+cos⁡y2f(x, y) = x e^{xy^2} + \cos y^2f(x,y)=xexy2+cosy2.

STEP 1

STEP 2

STEP 3

STEP 4

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

STEP 5

Calculate the derivative of xxx with respect to xxx.∂∂x(x)=1\frac{\partial}{\partial x} (x) = 1∂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)∂x∂​(xexy2)=xy2exy2+exy2(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}∂x∂​(xexy2)=xy2exy2+exy2

STEP 8

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

STEP 9

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

STEP 10

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

STEP 11

STEP 12

STEP 13

Since xxx is a constant with respect to yyy, its derivative with respect to yyy is zero.∂∂y(x)=0\frac{\partial}{\partial y} (x) = 0∂y∂​(x)=0

STEP 14

Calculate the derivative of exy2e^{x y^2}exy2 with respect to yyy 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}∂y∂​(exy2)=∂y∂​(xy2)exy2

STEP 15

Calculate the derivative of xy2x y^2xy2 with respect to yyy.∂∂y(xy2)=2xy\frac{\partial}{\partial y} \left( x y^2 \right) = 2xy∂y∂​(xy2)=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}∂y∂​(exy2)=2xyexy2

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)∂y∂​(xexy2)=x(2xyexy2)+exy2(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}∂y∂​(xexy2)=2x2yexy2

STEP 19

Calculate the partial derivative of the second term with respect to yyy 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)∂y∂​(cos(y2))=−sin(y2)∂y∂​(y2)

STEP 20

Calculate the derivative of y2y^2y2 with respect to yyy.∂∂y(y2)=2y\frac{\partial}{\partial y} (y^2) = 2y∂y∂​(y2)=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)∂y∂​(cos(y2))=−sin(y2)(2y)

STEP 22

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

STEP 23

Combine the derivatives from steps 18 and 22 to find the partial derivative of f(x,y)f(x, y)f(x,y) with respect to yyy.∂f∂y=2x2yexy2−2ysin⁡(y2)\frac{\partial f}{\partial y} = 2x^2y e^{x y^2} - 2y \sin(y^2)∂y∂f​=2x2yexy2−2ysin(y2)

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).∇f(x,y)=(∂f∂x,∂f∂y)\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)∇f(x,y)=(∂x∂f​,∂y∂f​)

STEP 25

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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