Question

If F(x,y) = x^2sin(xy), find Fyx.

Answer 1

`F_(yx)=3x^(2) cos(xy)- yx^(3) sin(xy)`

Explanation:

We need to find out the partial differential
`F_(yx)` of
`F(x,y)=x^(2)sin(xy)`

First, differentiate
`F(x,y)=x^(2)sin(xy)` both the sides with respect to 'y'

`(d)/(dy)F(x,y)=(d)/(dy)x^(2)sin(xy)`

Since,
`(d)/(dt)\sin t =\cos t`

`(d)/(dy)F(x,y)=x^(2)cos(xy)* (d)/(dy)(xy)`

`(d)/(dy)F(x,y)=x^(2)cos(xy)* x`

`(d)/(dy)F(x,y)=x^(3)cos(xy)`

so,
`F_y=x^(3)cos(xy)`

Now, differentiate above both the sides with respect to 'x'

`F_(yx)=(d)/(dx)x^(3)cos(xy)`

Chain rule of differentiation:
`D(fg)=f'g + fg'`

`F_(yx)=cos(xy) (d)/(dx)x^(3) + x^(3) (d)/(dx)cos(xy)`

Since,
`(d)/(dx)x^(m) =mx^(m-1)` and
`(d)/(dt) cost =-\sin t`

`F_(yx)=cos(xy)* 3x^(2) - x^(3) sin(xy)* (d)/(dx)(xy)`

`F_(yx)=cos(xy)* 3x^(2) - x^(3) sin(xy)* y`

`F_(yx)=3x^(2) cos(xy)- yx^(3) sin(xy)`

hence,
`F_(yx)=3x^(2) cos(xy)- yx^(3) sin(xy)`