find the total differential of the function
f(x,y)= xye^(x+y)
Well, to find the total differential of the function f(x, y) = xye^(x+y), we need to differentiate it with respect to x and y.
Let's start with differentiating with respect to x:
∂f/∂x = ∂/∂x (xye^(x+y))
Using the product rule, we get:
∂f/∂x = y * e^(x+y) + xye^(x+y)
Now, differentiating with respect to y:
∂f/∂y = ∂/∂y (xye^(x+y))
Again, using the product rule, we get:
∂f/∂y = x * e^(x+y) + xye^(x+y)
So, the total differential is:
df = (∂f/∂x)dx + (∂f/∂y)dy
= [y * e^(x+y) + xy * e^(x+y)]dx + [x * e^(x+y) + xy * e^(x+y)]dy
Well, that was a mouthful! But hey, at least we got to use the product rule twice. Gotta love those math shenanigans!
To find the total differential of the function f(x, y) = xye^(x+y), we need to take partial derivatives with respect to x and y.
First, let's find the partial derivative with respect to x:
∂f/∂x = ∂/∂x (xye^(x+y))
= y(e^(x+y) + xye^(x+y))
Next, let's find the partial derivative with respect to y:
∂f/∂y = ∂/∂y (xye^(x+y))
= x(e^(x+y) + xye^(x+y))
Now, we can write the total differential of f as:
df = (∂f/∂x)dx + (∂f/∂y)dy
Substituting the partial derivatives into the equation, we get:
df = [y(e^(x+y) + xye^(x+y))]dx + [x(e^(x+y) + xye^(x+y))]dy
Therefore, the total differential of f(x, y) = xye^(x+y) is:
df = [y(e^(x+y) + xye^(x+y))]dx + [x(e^(x+y) + xye^(x+y))]dy
To find the total differential of the function f(x, y) = xye^(x+y), we need to take the partial derivatives with respect to both x and y and then multiply them with the corresponding differentials dx and dy. The total differential can be written as follows:
df = (∂f/∂x)dx + (∂f/∂y)dy
Let's calculate the partial derivatives first:
∂f/∂x = y * e^(x+y) + xy * e^(x+y) ⟶ ∂f/∂x = (y + xy) * e^(x+y)
∂f/∂y = x * e^(x+y) + xy * e^(x+y) ⟶ ∂f/∂y = (x + xy) * e^(x+y)
Now, substitute these partial derivatives into the total differential formula:
df = (y + xy) * e^(x+y) * dx + (x + xy) * e^(x+y) * dy
So, the total differential of f(x, y) = xye^(x+y) is:
df = (y + xy) * e^(x+y) * dx + (x + xy) * e^(x+y) * dy
df = ∂f/∂x dx + ∂f/∂y dy
f = xye^(x+y)
df = (ye^(x+y) + xye^(x+y)) dx + (xe^(x+y) + xye^(x+y)) dy
= e^(x+y) [ (x+1)y dx + (y+1)x dy]
or, separating the product a bit,
f = xe^x * ye^y
df = e^x(x+1)*ye^y dx + e^y(y+1)*xe^x dy
= e^x*e^y [(x+1)y dx + (y+1)x dy]