Calculate all four second-order partial derivatives of f(x,y)=sin(5x/y)
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fxx(x,y)=
fxy(x,y)=
fyx(x,y)=
fyy(x,y)=
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)
To calculate the four second-order partial derivatives of f(x,y) = sin(5x/y), we need to differentiate the function twice with respect to each variable. Let's calculate them one by one:
1. fxx(x,y):
To find the second partial derivative with respect to x, we need to differentiate sin(5x/y) twice with respect to x.
First, let's find the first derivative:
d/dx [sin(5x/y)] = (5/y) * cos(5x/y)
Now, let's find the second derivative by differentiating the above result with respect to x:
d/dx[(5/y) * cos(5x/y)] = -[(5/y)^2] * sin(5x/y)
Therefore, fxx(x,y) = -[(5/y)^2] * sin(5x/y)
2. fxy(x,y):
To find the mixed partial derivative with respect to x and y, we need to differentiate sin(5x/y) with respect to x first and then with respect to y.
First, let's find the first derivative with respect to x:
d/dx[sin(5x/y)] = (5/y) * cos(5x/y)
Now, let's differentiate the above result with respect to y:
d/dy[(5/y) * cos(5x/y)] = -(5/y^2) * cos(5x/y)
Therefore, fxy(x,y) = -(5/y^2) * cos(5x/y)
3. fyx(x,y):
To find the mixed partial derivative with respect to y and x, we need to differentiate sin(5x/y) with respect to y first and then with respect to x.
First, let's find the first derivative with respect to y:
d/dy[sin(5x/y)] = -(5x/y^2) * cos(5x/y)
Now, let's differentiate the above result with respect to x:
d/dx[-(5x/y^2) * cos(5x/y)] = (-5/y^2) * cos(5x/y) - (25x/y^3) * sin(5x/y)
Therefore, fyx(x,y) = (-5/y^2) * cos(5x/y) - (25x/y^3) * sin(5x/y)
4. fyy(x,y):
To find the second partial derivative with respect to y, we need to differentiate sin(5x/y) twice with respect to y.
First, let's find the first derivative with respect to y:
d/dy[sin(5x/y)] = -(5x/y^2) * cos(5x/y)
Now, let's find the second derivative by differentiating the above result with respect to y:
d/dy[-(5x/y^2) * cos(5x/y)] = (10x/y^3) * cos(5x/y) + (25x^2/y^4) * sin(5x/y)
Therefore, fyy(x,y) = (10x/y^3) * cos(5x/y) + (25x^2/y^4) * sin(5x/y)
To summarize:
fxx(x,y) = -[(5/y)^2] * sin(5x/y)
fxy(x,y) = -(5/y^2) * cos(5x/y)
fyx(x,y) = (-5/y^2) * cos(5x/y) - (25x/y^3) * sin(5x/y)
fyy(x,y) = (10x/y^3) * cos(5x/y) + (25x^2/y^4) * sin(5x/y)