Question

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

Answer 1

Answer:

$F_{yx}=3x^{2} cos(xy)- yx^{3} sin(xy)$

Step-by-step 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'

$\frac{d}{dy}F(x,y)=\frac{d}{dy}x^{2}sin(xy)$

Since, $\frac{d}{dt}\sin t =\cos t$

$\frac{d}{dy}F(x,y)=x^{2}cos(xy)\times \frac{d}{dy}(xy)$

$\frac{d}{dy}F(x,y)=x^{2}cos(xy)\times x$

$\frac{d}{dy}F(x,y)=x^{3}cos(xy)$

so, $F_y=x^{3}cos(xy)$

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

$F_{yx}=\frac{d}{dx}x^{3}cos(xy)$

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

$F_{yx}=cos(xy) \frac{d}{dx}x^{3} + x^{3} \frac{d}{dx}cos(xy)$

Since, $\frac{d}{dx}x^{m} =mx^{m-1}$ and $\frac{d}{dt} cost =-\sin t$

$F_{yx}=cos(xy)\times 3x^{2} - x^{3} sin(xy)\times \frac{d}{dx}(xy)$

$F_{yx}=cos(xy)\times 3x^{2} - x^{3} sin(xy)\times y$

$F_{yx}=3x^{2} cos(xy)- yx^{3} sin(xy)$

hence, $F_{yx}=3x^{2} cos(xy)- yx^{3} sin(xy)$