Calculate all four second-order partial derivatives of f(x,y)=sin(5x/y)

Question

.

fxx(x,y)=


fxy(x,y)=


fyx(x,y)=


fyy(x,y)=

Answer 1

To calculate the second-order partial derivatives of f(x,y)=sin(5x/y), we will differentiate the function twice with respect to each variable.

First, we find the first-order partial derivatives:

fx(x,y) = d(sin(5x/y))/dx
= 5/y * cos(5x/y)

fy(x,y) = d(sin(5x/y))/dy
= -5x/y^2 * cos(5x/y)

Now, let's find the second-order derivatives:

fxx(x,y) = d^2(sin(5x/y))/dx^2
= d/dx (5/y * cos(5x/y))
= -25/y^2 * cos(5x/y) - 25/y^2 * sin(5x/y) * (-x/y)
= -25/y^2 * cos(5x/y) + 25x/y^3 * sin(5x/y)

fxy(x,y) = d^2(sin(5x/y))/dxdy
= d/dy (5/y * cos(5x/y))
= -5/y^2 * cos(5x/y) - 5/y^2 * sin(5x/y) * (-x/y)
= -5/y^2 * cos(5x/y) + 5x/y^3 * sin(5x/y)

fyx(x,y) = d^2(sin(5x/y))/dydx
= d/dx (-5x/y^2 * cos(5x/y))
= -(-5/y^2 * cos(5x/y) + 5x/y^3 * sin(5x/y))
= 5/y^2 * cos(5x/y) - 5x/y^3 * sin(5x/y)

fyy(x,y) = d^2(sin(5x/y))/dy^2
= d/dy (-5x/y^2 * cos(5x/y))
= 10x/y^3 * cos(5x/y) + 5x^2/y^4 * sin(5x/y)
= 10x/y^3 * cos(5x/y) + 5x^2/y^4 * sin(5x/y)

Therefore, the four second-order partial derivatives of f(x,y)=sin(5x/y) are:

fxx(x,y) = -25/y^2 * cos(5x/y) + 25x/y^3 * sin(5x/y)

fxy(x,y) = -5/y^2 * cos(5x/y) + 5x/y^3 * sin(5x/y)

fyx(x,y) = 5/y^2 * cos(5x/y) - 5x/y^3 * sin(5x/y)

fyy(x,y) = 10x/y^3 * cos(5x/y) + 5x^2/y^4 * sin(5x/y)