MATEMATİK I - Deneme Sınavı - 2

correct answer is E.

Thus:

f(x,y)=(x²-y²)/(5x-5y)=(x-y)(x+y)/5(x-y)=(x+y)/5=2/5 for x=1 and y=1. So the answer is B.

Then we can rewrite the function as z=e^xy?

Taking the natural logarithm of both sides we have:

lnz=lne^xy=xy

Taking the derivatives of the both sides we have:

(1/z)dz=ydx+xdy

When we take the partial derivative we behave all the other variables as constant. Thus here dy=0. Therefore we can rewrite the last expression as:

(1/z)dz=ydx

Thus dz/dx=yz=ye^xy since z=e^xy

Assume that z=f(x,y)

So z=e^xy

lnz=lne^xy=xy

Taking derivatives of both sides we get:

(1/z)dz=xdy+ydx and for dz/dx we consider dy=0

Thus dz/dx=yz=ye^xy

Let's assume that t=dz/dx. We now have to find dt/dy according to chain rule. So:

t=ye^xy?

dt=e^xydy+yxe^xydy (from the rule if a=bc then a'=cb'+bc')

So:

dt=(1+xy)e^xydy

dt/dy=(1+xy)e^xy

So the answer is A.

Thus for f (x, y) = x²+3y² :

f_x =2x=2 for x=1

f_y =6y=6 for y=1

and z=f(x,y)=4 for (x,y)=(1,1)

The equation for the tangent plane is given by z=f(a,b)+f_x(x-a)+f_y(y-b)

Therefore the equation for the tangent plane for a=1, b=1 is given by

z=4+2(x-1)+6(y-1)=2x+6y-4

Thus for f (x, y) = x²+3y² :

f_x =2x=2 for x=1

f_y =6y=6 for y=1

and z=f(x,y)=4 for (x,y)=(1,1)

The equation for the tangent line is given by z=f(a,b)+f_x(x-a)+f_y(y-b)

Therefore the equation for the tangent plane for a=1, b=1 is given by

z=4+2(x-1)+6(y-1)=2x+6y-4

therefore the equation for the normal line is

(z-4)/-1=(x-1)/2+(y-1)/6

so 4-z=(x-1)/2+(y-1)/6

which means:

24-6z=3x-3=y-1

Given that z=x²+xy where x=u²v and y=2uv:

dz/dx=2x+y

dx/du=2vu

dz/dy=x

dy/du=2v

So for u=1 and v=2; x=u²v=2 and y=2uv=4

dz/dx=2x+y=2u²v+2uv=4+4=8

dx/du=2vu=4

dz/dy=x=2

dy/du=2v=4

Finally:

dz/du=(dz/dx)*(dx/du)+(dz/dy)*(dy/du)=8*4+2*4=40

Since f(x,y)=x²y²-x²-y²?

f_x=2xy²-2x=0 so 2x(y²-1)=0 which means either x=0 or y=1 or y=-1

f_y=2yx²-2y=0 so 2y(x²-1)=0 which means either y=0 or x=1 or x=-1

But since both equations must be satisfied simultaneously we have 5 critical points:

(0,0), (-1,1), (1,-1), (1,1) and (-1,-1).

As you can see the point (1,0) (namely x=1 and y=0) satisfies the second equation (fy=0) but it does not satisfy the first one. Thus the answer is D.