Q:

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

Accepted Solution

A:
Answer:[tex]F_{yx}=3x^{2} cos(xy)- yx^{3} sin(xy)[/tex]Step-by-step explanation:We need to find out the partial differential [tex]F_{yx}[/tex] of [tex]F(x,y)=x^{2}sin(xy)[/tex]First, differentiate [tex]F(x,y)=x^{2}sin(xy)[/tex] both the sides with respect to 'y'[tex]\frac{d}{dy}F(x,y)=\frac{d}{dy}x^{2}sin(xy)[/tex]Since, [tex]\frac{d}{dt}\sin t =\cos t[/tex][tex]\frac{d}{dy}F(x,y)=x^{2}cos(xy)\times \frac{d}{dy}(xy)[/tex][tex]\frac{d}{dy}F(x,y)=x^{2}cos(xy)\times x[/tex][tex]\frac{d}{dy}F(x,y)=x^{3}cos(xy)[/tex]so, [tex]F_y=x^{3}cos(xy)[/tex]Now, differentiate above both the sides with respect to 'x'[tex]F_{yx}=\frac{d}{dx}x^{3}cos(xy)[/tex]Chain rule of differentiation: [tex]D(fg)=f'g + fg'[/tex][tex]F_{yx}=cos(xy) \frac{d}{dx}x^{3} + x^{3} \frac{d}{dx}cos(xy)[/tex]Since, [tex] \frac{d}{dx}x^{m} =mx^{m-1}[/tex] and [tex] \frac{d}{dt} cost =-\sin t[/tex][tex]F_{yx}=cos(xy)\times 3x^{2} - x^{3} sin(xy)\times \frac{d}{dx}(xy)[/tex][tex]F_{yx}=cos(xy)\times 3x^{2} - x^{3} sin(xy)\times y[/tex][tex]F_{yx}=3x^{2} cos(xy)- yx^{3} sin(xy)[/tex]hence, [tex]F_{yx}=3x^{2} cos(xy)- yx^{3} sin(xy)[/tex]