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

Studdy Solution

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:
$\nabla f(x, y) = \left( e^{x y^2} (x y^2 + 1), 2x^2y e^{x y^2} - 2y \sin(y^2) \right)$

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