For each function, hold every variable except the one being differentiated constant, apply the ordinary derivative rules, then substitute the given point.
Part (i): f(x,y)=3x2−2xy+y2+5x+2 at (2,−5).
fx=6x−2y+5. At (2,−5): 6(2)−2(−5)+5=12+10+5=27.
fy=−2x+2y. At (2,−5): −2(2)+2(−5)=−4−10=−14.
Part (ii): g(x,y)=3x2+y2+5x+2 at (1,−2).
gx=6x+5. At (1,−2): 6(1)+5=11.
gy=2y. At (1,−2): 2(−2)=−4.
Part (iii): h(x,y,z)=xsin(xy)+z2x at (2,4π,1).
hx=sin(xy)+x⋅cos(xy)⋅y+z2=sin(xy)+xycos(xy)+z2 (product rule on xsin(xy), plus z2 from z2x).
At the point: xy=2⋅4π=2π, so sin(xy)=1, cos(xy)=0. hx=1+(2π)(0)+12=1+0+1=2.
hy=x⋅cos(xy)⋅x=x2cos(xy) (only the xsin(xy) term depends on y). At the point: x2=4, cos(xy)=0, so hy=0.
hz=2zx (only z2x depends on z). At the point: 2(1)(2)=4.
Part (iv): G(x,y)=ex+3ylog(x2+y2) at (−1,1).
Gx=ex+3ylog(x2+y2)+ex+3y⋅x2+y22x=ex+3y[log(x2+y2)+x2+y22x] (product rule).
At (−1,1): x+3y=−1+3=2; x2+y2=1+1=2; x2+y22x=2−2=−1. So Gx=e2[log2−1].
Gy=3ex+3ylog(x2+y2)+ex+3y⋅x2+y22y=ex+3y[3log(x2+y2)+x2+y22y].
At (−1,1): x2+y22y=22=1. So Gy=e2[3log2+1].
✓Final answer
(i) fx(2,−5)=27, fy(2,−5)=−14 (ii) gx(1,−2)=11, gy(1,−2)=−4 (iii) hx=2, hy=0, hz=4 (iv) Gx=e2(log2−1)≈−1.29, Gy=e2(3log2+1)≈22.75