Find the gradient of xy^2 + x^2y=4xy at the point (1,3)?
If you mean dy/dx, then
D( xy^2 + x^2y)=D(4xy)
y^2+2yy'+2xy+x^2 y'=4y+4xy'
y'(2y+x^2-4x)=4y+y^2+2xy
so at point 1,3
y'(6+1-4)=12+9+6
y'=37/3
check that.
Find the gradient of the tangent line of x^2y+xy^2=xy at (-1;2)
To find the gradient of the function xy^2 + x^2y = 4xy at the point (1,3), we need to find the partial derivatives of the function with respect to x and y.
Step 1: Find ∂(xy^2 + x^2y)/∂x:
Taking the derivative of the function with respect to x while treating y as a constant:
∂(xy^2 + x^2y)/∂x = y^2 + 2xy
Step 2: Find ∂(xy^2 + x^2y)/∂y:
Taking the derivative of the function with respect to y while treating x as a constant:
∂(xy^2 + x^2y)/∂y = 2xy + x^2
Step 3: Evaluate at the point (1,3):
Substitute x = 1 and y = 3 into the partial derivatives obtained in steps 1 and 2:
∂(xy^2 + x^2y)/∂x at (1,3) = 3^2 + 2(1)(3) = 9 + 6 = 15
∂(xy^2 + x^2y)/∂y at (1,3) = 2(1)(3) + 1^2 = 6 + 1 = 7
Step 4: The gradient vector is given by (∂f/∂x, ∂f/∂y), where f is the function.
The gradient at the point (1,3) is (15, 7).
To find the gradient of the function xy^2 + x^2y = 4xy at the point (1,3), you can use partial derivatives.
The gradient of a function of two variables is a vector that points in the direction of the steepest increase in the function and has a magnitude equal to the rate of change of the function in that direction.
To find the gradient, you need to compute the partial derivatives of the function with respect to each variable, x and y. These partial derivatives will give you the rates of change in each direction.
Step-by-step Solution:
1. Start by taking the partial derivative with respect to x. Treat y as a constant while differentiating.
- Differentiating xy^2 with respect to x gives y^2.
- Differentiating x^2y with respect to x gives 2xy.
- Differentiating 4xy with respect to x gives 4y.
2. Now, take the partial derivative with respect to y. Treat x as a constant while differentiating.
- Differentiating xy^2 with respect to y gives 2xy.
- Differentiating x^2y with respect to y gives x^2.
- Differentiating 4xy with respect to y gives 4x.
3. Combine the partial derivatives to form the gradient vector.
- Gradient vector = [2xy + 2xy, y^2 + x^2, 4x + 4y]
- Simplify the vector to [4xy, y^2 + x^2, 4x + 4y]
4. Evaluate the gradient vector at the point (1,3).
- Plug in x = 1 and y = 3 into the gradient vector.
- Gradient at (1,3) = [4(1)(3), (3^2) + (1^2), 4(1) + 4(3)]
= [12, 9 + 1, 4 + 12]
= [12, 10, 16]
Therefore, the gradient of the function xy^2 + x^2y = 4xy at the point (1,3) is [12, 10, 16].