Touba M. answered • 11/22/21
B.S. in Pure Math with 20+ Years Teaching/Tutoring Experience
hi,
for finding derivative, please listen to me
(2x)' = 2
(ax)' = a
(yx) = y agree with me? if y has imagined a number, yes it is true!
1) f(x,y)= y^2+y(sin(x)-x^4)
f'(x,y) = - f'x / f'y ===============> it means: - f'x you take a derivative of all terms that include x and imagine y is a number SO [y(sin(x)-x^4)]' = ycos(x) -4x3
THEN f'y it means you take of derivative of all terms that include y and imagine x is a number
SO [y^2+y(sin(x)]' = 2y + sin(x)
NOW f'(x,y) = - f'x / f'y = - [ ycos(x) -4x3 ] / [ 2y + sin(x) ]
f(x,y)= xy+2x-ln(x^2y) -------> f'(x,y) = - [ y + 2 - (2xy/x^2y)] / [ x - (x^2) / (x^2y)] don't forget to simplify
f(x,y)= x^2e^(xy) --------------> f'(x,y) =-[2x e^(xy) + y e^(xy) * x2] / [ x e^(xy)* x2 ]
I hope it works for you,
Minoo
Touba M.
11/23/21
Luke J.
Touba, why are the solutions being represented as a negative ratio of one over the other? I'm confused as to why the solution would be shown this way. Partial derivatives do not have to be inter-related nor do they coincide with an overall derivative of the first multivariable function f. I believe this posted solution deserves some more space and clarity in the direction of where it is supposed to go. None of your definitions of partial derivatives are incorrect, they are spot on but the execution thereafter is considerably vague and not the original point of the problem. Each function would have 2 separate, pretty nearly unrelated partial derivatives since they are functions of x and y thus would have partial f over partial x or partial f over partial y equal to their respective equations. I'd urge you to consider a revision of this problem for the peer who posted this problem and for the clarity of anyone else who looks at this problem.11/22/21