Tanget plane of z=ycos(x-y), at point (2,2,2)

[day1ok]

Find the tanget plane using partial derivatives.

z=ycos(x-y), at point (2,2,2)


attempted: fx(x,y)=ysin(x-y)*1
fx(2,2)=0
fy(x,y)=1*sin(x-y)*1
fx(2,2)=0
z-2=0(x-2)+0(x-2)
z=2 Incorrect

 

[Hootenanny]

Find the tanget plane using partial derivatives.

z=ycos(x-y), at point (2,2,2)attempted: fx(x,y)=ysin(x-y)*1
fx(2,2)=0
fy(x,y)=1*sin(x-y)*1
fx(2,2)=0
z-2=0(x-2)+0(x-2)
z=2 Incorrect

Welcome to Physics Forums.

Are you sure about your evaluation of f_x and f_y?

 

[day1ok]

no I am noy sure i may be finding the partial derivative of both fx and fy wrong

 

[gabbagabbahey]

You need to use the product rule to find fy since there is a y both in front of the sin, and in its argument.

 

[day1ok]

got it thanks