a)find the first partial derivatives of f(x y)= x √1+y^2
b)find the first partial derivatives of f(x,y)= e^x ln y at the point (0,e)
f(x,y) = x√(1+y^2)
fx = (√1+y^2)
fy = xy/(√1+y^2)
Just treat the other variables as constants.
fx=e^xlny
fy=e^x/y
fx=e^(0)ln(e)=1
fy=1/e
a) To find the first partial derivatives of a two-variable function f(x, y), you have to differentiate the function with respect to each variable separately. In this case, let's find the two partial derivatives of f(x, y) = x √(1 + y^2).
Partial derivative with respect to x (denoted as ∂f/∂x):
To calculate this, treat y as a constant and differentiate x with respect to x, considering that the derivative of x with respect to x is 1. So, ∂f/∂x = √(1 + y^2).
Partial derivative with respect to y (denoted as ∂f/∂y):
To calculate this, treat x as a constant and differentiate √(1 + y^2) with respect to y. We need to use the chain rule here. The derivative of √(1 + y^2) with respect to y is 2y/2√(1 + y^2), but since we are calculating the partial derivative with respect to y, we multiply it by the derivative of the expression inside the square root, which is 2y. So, ∂f/∂y = x * 2y/2√(1 + y^2) = xy/√(1 + y^2).
Therefore, the first partial derivatives of f(x, y) = x √(1 + y^2) are:
∂f/∂x = √(1 + y^2)
∂f/∂y = xy/√(1 + y^2)
b) To find the first partial derivatives of the function f(x, y) = e^x ln y at the point (0, e):
Partial derivative with respect to x (denoted as ∂f/∂x):
To calculate this, treat y as a constant and differentiate e^x with respect to x. The derivative of e^x with respect to x is simply e^x. So, ∂f/∂x = e^x ln y.
Partial derivative with respect to y (denoted as ∂f/∂y):
To calculate this, treat x as a constant and differentiate ln y with respect to y. The derivative of ln y with respect to y is simply 1/y. So, ∂f/∂y = e^x * 1/y = e^x/y.
Therefore, the first partial derivatives of f(x, y) = e^x ln y at the point (0, e) are:
∂f/∂x = e^0 ln e = ln e = 1
∂f/∂y = e^0 / e = 1/e