Find the mixed derivative f _(x y ) of f(x ,y)=y e ^(x^2+y^2)?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need first to differentiate the function with respect to x, considering y as constant, then you need to differentiate the partial derivative with respect to y, considering x as constant.

`(del f)/(del x) = (del(ye^(x^2+y^2)))/(del x)`

`(del f)/(del x) = y*(del e^(x^2+y^2))/(del x)`

`(del f)/(del x) = y*e^(x^2+y^2)*(del(x^2 + y^2))/(del x)`

`(del f)/(del x) = y*e^(x^2+y^2)*(2x^2 + 0)`

`(del f)/(del x) = 2x^2*y*e^(x^2+y^2)`

You need to differentiate `(del f)/(del x)` with respect to y, using product rule, such that:

`(del((del f)/(del x)))/(del y) = (del 2x^2*y)/(del y)*e^(x^2+y^2) +2x^2*y* (del e^(x^2+y^2))/(del y)`

`(del((del f)/(del x)))/(del y) = 2x^2*e^(x^2+y^2) + 2x^2*y*e^(x^2+y^2)*2y`

`(del((del f)/(del x)))/(del y) = 2x^2*e^(x^2+y^2) +4x^2*y^2*e^(x^2+y^2)`

Factoring out `2x^2*e^(x^2+y^2)` yields:

`f_(x,y) = (del((del f)/(del x)))/(del y) = 2x^2*e^(x^2+y^2)(1 + 2y^2)`

Hence, evaluating the mixed derivative `f_(x,y)` , yields,` f_(x,y) = 2x^2*e^(x^2+y^2)(1 + 2y^2).`

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