What is the dz/dx and dz/dy of

Question

z= [3e^-((2x^2 + 3y^2)/10)] + sin(x^2 + 2y^2)?  

 

 

Answer 1

Derivate with respect to x.

Derivate with respect to y.

Therefore dz/dx = -6x/5 e^{-(2x^2 + 3y^2)/10} + 2xcos(x² + 2y²) and

dz/dy = -9y/5 e^{-(2x^2 + 3y^2)/10} + 4ycos(x² + 2y²)