Answer to Question #107589 in Calculus for jennifer

Question

Given that z= xysin(2x+3y) find (a) z_xx (b) z_yy

Answer 1

(a) $z=xysin(2x+3y)$

differentiating with respect to x considering y as constant (using product rule)

$z_x=xy\cdot cos(2x+3y)\cdot2+sin(2x+3y)\cdot y$

$z_x=2xycos(2x+3y)+ysin(2x+3y)$

differentiating with respect to x once again

$z_{xx}=2xy\cdot (-sin(2x+3y))\cdot2+cos(2x+3y)\cdot2y+y\cdot cos(2x+3y)\cdot2$

$z_{xx}=-4xysin(2x+3y)+4ycos(2x+3y)$

(b) $z=xysin(2x+2y)$

differentiating with respect to y considering x as constant ( using product rule)

$z_y=xy\cdot cos(2x+3y)\cdot3+sin(2x+3y)\cdot x$

$z_y=3xycos(2x+3y)+xsin(2x+3y)$

differentiating with respect to y again

$z_{yy}=3xy \cdot(-sin(2x+3y)\cdot 3)+cos(2x+3y)\cdot3x+x\cdot cos(2x+3y)\cdot 3$

$z_{yy}=-9xysin(2x+3y)+6xcos(2x+3y)$