Find the gradient of f(x,y). f(x,y)=(x^2)e^-2y

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The gradient of the function f(x,y)=(x^2)e^-2y is definitively calculated as grad f = <2xe^-2y, -4xe^-2y>. The correct partial derivatives are fx(x,y)=(2x)e^-2y and fy(x,y)=-2x^2e^-2y. The initial confusion stemmed from misinterpretations of the derivatives, particularly in handling the exponential function's derivative with respect to y. This discussion clarifies the correct application of differentiation rules for multivariable functions.

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Here's the problem. Find the gradient of f(x,y). f(x,y)=(x^2)e^-2y.


I don't have the solution to this and I need to know if I got the right gradient (I have more problems that depend on this gradient, points on it). I ended up getting, gradient f=<2xe^-2y, 2x-2e^-2y>. I don't think it's right, but can someone help me out here?
 
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No.
grad f= fx(x,y)i + fy(x,y)j
fx(x,y)=(2x)e-2y
fy(x,y)=(-2*2x)e-2y
 
Sorry, Stephen, you have fy wrong.

The derivative of e-2y with respect to y is -2 e-2y The other factor, x2 is independent of y so treat it like a constant fy= (x2)(-2e-2y)= -2x2e-2y.

The gradient of 2xe-2y is the vector <2x e-2y, -4xe-2y>.

What ffrpg wrote: f=<2x^e-2y, 2x-2e^-2y> may be typos or just carelessness: x^e-2y doesn't make much sense and in "2x-2..." you MEANT (2x) times (-2), not 2x subtract 2...
 
Ah yes. Where is my head?
 

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