Use logarithmic differentiation to find y ′ for the following function.
x^3 = e^−4y4−9y
I assume you mean
x^3 = e^(−4y^4−9y)
otherwise using logs makes no sense. So, we have
3logx = -4y^4-9y
3/x dx/dy = -16y^3-9
dx/dy = -x/3 (16y^3+9)
so, dy/dx = -3/[x(16y^3+9)]
Not the usual formulation, eh?
using implicit differentiation, we get
3x^2 = -(16y^3+9)e^(−4y^4−9y) y'
y' = -3x^2/[e^(−4y^4−9y)(16y^3+9)]
But that bottom factor of e^(−4y^4−9y) is just x^3, so you again have
y' = -3/[x(16y^3+9)]