Find all first partial derivatives:
f(x,y)=e^(xy)cosx
Fx = ye^(xy)cosx - e^(xy)sinx
= e^(xy) (ycosx - sinx)
Fy = xe^(xy) cosx
Find, f1(0,180°) if (x,y)= e^(xy)cos(x+y).
Find, f1(0,180°) if f(x,y)=e^(xy)cos(x+y)
To find all the first partial derivatives of the function f(x, y) = e^(xy)cosx, we need to differentiate with respect to each variable separately.
Let's start by taking the partial derivative with respect to x, denoted as ∂/∂x.
Step 1: Differentiate e^(xy) with respect to x.
The derivative of e^(xy) with respect to x is ye^(xy), using the chain rule.
Step 2: Differentiate cosx with respect to x.
The derivative of cosx with respect to x is -sinx.
Combining the results from Step 1 and Step 2, we have:
∂f/∂x = ye^(xy)cosx - sinx
Now, let's find the partial derivative with respect to y, denoted as ∂/∂y.
Step 1: Differentiate e^(xy) with respect to y.
The derivative of e^(xy) with respect to y is x * e^(xy), again using the chain rule.
Step 2: Differentiate cosx with respect to y.
Since the variable y does not appear in cosx, the derivative with respect to y is 0.
Combining the results from Step 1 and Step 2, we have:
∂f/∂y = x * e^(xy)
So, the first partial derivatives are:
∂f/∂x = ye^(xy)cosx - sinx
∂f/∂y = x * e^(xy)