What are the partial derivatives of cos(e^x+y) ?
1 Answer
If
# f_x = e^xcos(e^x+y) #
# f_y = cos(e^x+y) #
Whereas, if:
# f(x,y) = sin(e^(x+y)) #
Then:
# f_x = e^(x+y)cos(e^(x+y)) #
# f_y = e^(x+y)cos(e^(x+y)) #
Explanation:
I am non sure if your mean
# f(x,y) = sin(e^x+y) # , or
# f(x,y) = sin(e^(x+y)) #
So I will consider both. Remember when partially differentiating that we differentiate wrt the variable in question whilst treating the other variables as constant. And so:
If:
# f(x,y) = sin(e^x+y) #
Then:
# f_x = (partial f)/(partial x) #
# \ \ \ = (partial)/(partial x) (sin(e^x+y)) #
# \ \ \ = cos(e^x+y) (partial)/(partial x) (e^x+y) #
# \ \ \ = cos(e^x+y) (e^x) #
# \ \ \ = e^xcos(e^x+y) #
# f_y = (partial f)/(partial y) #
# \ \ \ = (partial)/(partial y) (sin(e^x+y)) #
# \ \ \ = cos(e^x+y) (partial)/(partial y) (e^x+y) #
# \ \ \ = cos(e^x+y) (1) #
# \ \ \ = cos(e^x+y) #
Whereas, if:
# f(x,y) = sin(e^(x+y)) #
Then:
# f_x = (partial f)/(partial x) #
# \ \ \ = (partial)/(partial x) (sin(e^(x+y))) #
# \ \ \ = cos(e^(x+y)) (partial)/(partial x) (e^(x+y)) #
# \ \ \ = cos(e^(x+y)) (e^(x+y)) #
# \ \ \ = e^(x+y)cos(e^(x+y)) #
# f_y = (partial f)/(partial y) #
# \ \ \ = (partial)/(partial y) (sin(e^(x+y))) #
# \ \ \ = cos(e^(x+y)) (partial)/(partial y) (e^(x+y)) #
# \ \ \ = cos(e^(x+y)) (e^(x+y)) #
# \ \ \ = e^(x+y)cos(e^(x+y)) #
