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))
