(a) At a glance this is wrong because of:
A. The answer is not a linear function;
B. The partial derivatives were not evaluated at the point.
(b) Common equation for the tangent plane:
Fx′(x0,y0,z0)(x−x0)+Fy′(x0,y0,z0)(y−y0)+
+Fz′(x0,y0,z0)(z−z0)=0 , then:
z0=x04−y05=(4)4−(1)5=256−1=255;
Find the equation of the tangent plane to the surface z=x4−y5 at the point (x0,y0,z0)=(4,1,255):
F(x,y,z)=−z+x4−y5=0 ;
Fx′=4x3 ; Fx′(x0,y0,z0)=4⋅43=256 ;
Fy′=−5y4 ; Fy′(x0,y0,z0)=−5⋅14=−5 ;
Fz′=−1 ; Fz′(x0,y0,z0)=−1 ;
256(x−x0)−5(y−y0)−(z−z0)=0 ;
256(x−4)−5(y−1)−(z−255)=0 ;
z=255+256(x−4)−5(y−1).
Answer: (a)A, B;
(b) z=255+256(x−4)−5(y−1).
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