# Find all second partial derivatives, fxx(x, y), fxy(x, y), fyx(x, y), and fyy(x, y)(a) f(x, y) = ln (5x^2 – 7y^3) and (b) x^2y^3e^(2x+3y)

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### 1 Answer

While finding partial derivatives of f(x,y) with respect x, we differentiate with respect x treating y as constant . Similarly x is treated to be like a constant , while finding partial derivative of y and f(x,y) is differentiated with respect to y.

f(x,y) =ln(5x^2-7y^3)

fx(x,y) = [1/(5x^2-7y^3)]10x

fxx(x,y) = [-1/(5x^2-7y^3)^2]100x^2+[1/(5x^2-7y^3]10x

fyx(x,y) = [-1/(5x^2-7y^3)^2]10x*(-7y^2)+0,as 10x is treated as constant with respect to y.

fy(x,y) = [1/(5x^2-7y^3)](-7y^2)

fxy(x,y) =[-1/(5x^2-7y^3)^2](-7y^2)(10x)+0 as -7y^2 is treated as constant with respect to x.

fyy(x,y) = [-1/(5x^2-7y^2)](-7y^2)^2 +[-1/(5x^2-7y^2)](-7*2*y)

b) f(x,y) = x^2y^3e^(2x+3y)=(x^2e^(2x))(3y^2e^(3y)

fx(x,y) = (2xe^(2x)+x^2e^2*2)3y^2e^(3y)=2x(x+1)y^3e^(2x+3y)

fxx(x,y) = (4x+2)y^3e^(2x+3y)+[x(x+1)]^2 *y^3*e^(2x+3y)

fyx(x,y) = 2x(x+1){3y^2e^(2x+3y)+y^3e^(2x+3y)*3} = 2x(x+1)*3y^2(y+1)e^(2x+3y)

fy(x,y) = x^2[3y^2e^(2x+3y)+y^3e^(2x+3y)*3] = =3x^2y^2(y+1)e^(2x+3y)

fxy(x,y) =3y^2(y+1) [x^2e^(2x+3y)*(2)+2xe^(2x+3y)] =

=3 y^2(y+1)(2x^2+2x) e^(2x+3y) = 6x(x+1)y^2(y+1)e^(2x+3y)

fyy(x,y) = 3x^2{y^2(y+1)e^(2+3y)*3+(3y^2+2y)e^(2x+3y)]

=3x^2*{3y^3+6y^2+2y}e^(2x+3y)