ANSWER : f. z=5(x−1)z=5(x-1)z=5(x−1)
EXPLANATION
The tangent plane to z=h(x,y) at the point (a,b) is z−h(a,b)=hx′(a,b) (x−a)+hy′(a,b)(y−b)z-h(a,b)={ h }_{ x }^{ ' }(a,b)\ (x-a)+{ h }_{ y }^{ ' }(a,b)(y-b)z−h(a,b)=hx′(a,b) (x−a)+hy′(a,b)(y−b)
a=1,b=5, h(a,b) =0a=1,\quad b=5,\ h(a,b)\ =0\quada=1,b=5, h(a,b) =0
hx′(x,y)=yx,{ h }_{ x }^{ ' }(x,y)=\frac { y }{ x } ,\quadhx′(x,y)=xy, hx′(1,5)=5hy′(x,y)=lnx,hy′(1,5)=0{ h }_{ x }^{ ' }(1,5)=5\quad { h }_{ y }^{ ' }(x,y)=\ln { x } , { h }_{ y }^{ ' }(1,5)=0hx′(1,5)=5hy′(x,y)=lnx,hy′(1,5)=0 .
So, the tangent plane is z=5(x−1)z=5(x-1)z=5(x−1)
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