f(x,y)=ln(5x^2-7y^3)
Find ALL 2nd derivatives.
fxx,fxy,fyx,fyy
To find the second derivatives of the function f(x, y) = ln(5x^2 - 7y^3), we need to take partial derivatives twice. Here's how you can do it step by step:
Step 1: Find the first partial derivatives.
The first partial derivative with respect to x (f_x) can be found by treating y as a constant and differentiating with respect to x:
f_x = d/dx [ln(5x^2 - 7y^3)]
= 10x / (5x^2 - 7y^3)
The first partial derivative with respect to y (f_y) can be found by treating x as a constant and differentiating with respect to y:
f_y = d/dy [ln(5x^2 - 7y^3)]
= -21y^2 / (5x^2 - 7y^3)
Step 2: Find the second partial derivatives.
To find the second partial derivatives, we need to differentiate the first partial derivatives obtained in Step 1.
The second partial derivative with respect to x (f_xx) can be found by differentiating f_x with respect to x:
f_xx = d/dx [f_x]
= d/dx [10x / (5x^2 - 7y^3)]
= (10(5x^2 - 7y^3) - 10x(10x)) / (5x^2 - 7y^3)^2
= (50x^2 - 70y^3 - 100x^2) / (5x^2 - 7y^3)^2
= (-50x^2 - 70y^3) / (5x^2 - 7y^3)^2
The second partial derivative with respect to y (f_yy) can be found by differentiating f_y with respect to y:
f_yy = d/dy [f_y]
= d/dy [-21y^2 / (5x^2 - 7y^3)]
= [42y(5x^2 - 7y^3) + 21y^2(21y^2)] / (5x^2 - 7y^3)^2
= (210xy - 294y^4 + 441y^4) / (5x^2 - 7y^3)^2
= (210xy + 147y^4) / (5x^2 - 7y^3)^2
The mixed second partial derivatives f_xy and f_yx can be found by differentiating f_x with respect to y and f_y with respect to x, respectively.
f_xy = d/dy [f_x]
= d/dy [10x / (5x^2 - 7y^3)]
= -210xy^2 / (5x^2 - 7y^3)^2
f_yx = d/dx [f_y]
= d/dx [-21y^2 / (5x^2 - 7y^3)]
= 210xy / (5x^2 - 7y^3)^2
So, the second derivatives are:
f_xx = (-50x^2 - 70y^3) / (5x^2 - 7y^3)^2
f_yy = (210xy + 147y^4) / (5x^2 - 7y^3)^2
f_xy = -210xy^2 / (5x^2 - 7y^3)^2
f_yx = 210xy / (5x^2 - 7y^3)^2