Derivatives and Integration | Calculus Help

Assignment   3

 

F(x,y)=x^2* y *sin(xy)

 

d/dx(F(x,y)) = x^(2)*y*sin(xy)

 

Multiply x^(2) by y to get x^(2)y.

x^(2)y*sin(xy)

 

Multiply x^(2)y by sin(xy) to get x^(2)ysin(xy).

x^(2)ysin(xy)

 

Find the derivative of the expression.

(d)/(dx) x^(2)ysin(xy)

 

Use the product rule to find the derivative of x^(2)ysin(xy).  The product rule states that (fg)’=f’g+fg’.

[(d)/(dx) x^(2)y](sin(xy))+(x^(2)y)[(d)/(dx) sin(xy)]

 

The derivative of x^(2)y is (d)/(dx) x^(2)y=2xy.

(d)/(dx) x^(2)y=2xy

 

Substitute the derivative back into the product rule formula.

(d)/(dx) x^(2)ysin(xy)=(2xy)(sin(xy))+(x^(2)y)[(d)/(dx) sin(xy)]

 

The derivative of sin(xy) is (d)/(dx) sin(xy)=ycos((xy)).

(d)/(dx) sin(xy)=ycos((xy))

 

Substitute the derivative back into the product rule formula.

(d)/(dx) x^(2)ysin(xy)=(2xy)(sin(xy))+(x^(2)y)(ycos((xy)))

 

Simplify the derivative.

(d)/(dx) x^(2)ysin(xy)=2xysin(xy)+x^(2)y^(2)cos(xy)

 

The derivative of x^(2)*y*sin(xy) is 2xysin(xy)+x^(2)y^(2)cos(xy).

2xysin(xy)+x^(2)y^(2)cos(xy)

 

 

ii.

 

And   d/dy(F(x,y)) = x^(2)*y*sin(xy)

 

Multiply x^(2) by y to get x^(2)y.

x^(2)y*sin(xy)

 

Multiply x^(2)y by sin(xy) to get x^(2)ysin(xy).

x^(2)ysin(xy)

 

Find the derivative of the expression.

(d)/(dy) x^(2)ysin(xy)

 

Use the product rule to find the derivative of x^(2)ysin(xy).  The product rule states that (fg)’=f’g+fg’.

[(d)/(dy) x^(2)y](sin(xy))+(x^(2)y)[(d)/(dy) sin(xy)]

 

The derivative of x^(2)y is (d)/(dy) x^(2)y=x^(2).

(d)/(dy) x^(2)y=x^(2)

 

Substitute the derivative back into the product rule formula.

(d)/(dy) x^(2)ysin(xy)=(x^(2))(sin(xy))+(x^(2)y)[(d)/(dy) sin(xy)]

 

The derivative of sin(xy) is (d)/(dy) sin(xy)=xcos((xy)).

(d)/(dy) sin(xy)=xcos((xy))

 

Substitute the derivative back into the product rule formula.

(d)/(dy) x^(2)ysin(xy)=(x^(2))(sin(xy))+(x^(2)y)(xcos((xy)))

 

Simplify the derivative.

(d)/(dy) x^(2)ysin(xy)=x^(2)sin(xy)+x^(3)ycos(xy)

 

The derivative of x^(2)*y*sin(xy) is x^(2)sin(xy)+x^(3)ycos(xy).

x^(2)sin(xy)+x^(3)ycos(xy)

 

 

So,

For the given functions the derivates are as follows

 

d/dx(f(x,y))= 2xysin(xy)+x^(2)y^(2)cos(xy)

d/dy(f(x,y))= x^(2)sin(xy)+x^(3)ycos(xy)

 

 

 

 

Z = x3 +3 xy^2-3x^2-3y^2+7

So now we will find Partial Derivates.

Fx =  3x^2+3y^2-6x   and Fy=6xy – 6y

To,

Find the Critical Points we solve Fx= 0 and Fy=0

3×2+3y2-6x=0   and  6xy-6y=0

Now from Fy=0

We get  6y=0  or   x  =  1

 

So,

When y = 0

We get  = 3×2 – 6x=0

X= 0  or  x  = 2

 

Or when  x= 1  we get

3 + 3y2-6=0

3y^2-3=0

Y= +-1

So,

 

The critical points are

(1,1) ,(1,-1),(0,0),(2,0)

 

To  identify nature  of  critical Points , we apply the second derivate test ,

 

A=6x- 6    B = 6y  C =6x-6

 

So  at the point  1,1   we have

A=0 , B=6, C=0

So  ,

Saddle point = (1,1) .

 

At the point (0,0)  we have

A= -6 , B=0 , C = -6

So,

(0,0) is  local maximum

 

At the point (2,0) we have

A= 6   B=0  C=6

AC-B^2

=  6 *6 – 0 ^2  = 36 >0

 

So,

(2,0) is  local maximum.

 

 

Given Z=x^2- xy+y^3- x

Let Z =f(x,y)=x^2 – xy +y^3 – x

 

Now we will find partial Derivative

Fx= 2x – y- 1   and Fy= -x+3y^2

 

To Find

Critical Points ,

We solve

2x-y- 1=0   -I

-x+3y^2=0  – I

 

X=3*y^2.

 

We put  value of x  in equation – I

2 * 3y^2 – y – 1 = 0

6*y^2 – y- 1 =0

 

Y= ½  and -1/3

 

Now if we put y=1/2  and  -1/3  in equation —ii

 

X = 3 * ¼    and   = 3  *1/9

= ¾  and  1/3.

 

So,

The critical  points are  (3/4,1/2)  and (1/3,-1/3)

 

To identify the nature of critical points , we apply the second derivate test , we have

A =fxx= 2    and  C =Fyy= 6y  and B=fxy=-1

 

At the point (3/4,1/2) we have , fxx= 2 , fxy=-1 ,fyy=3

Now  at given point ,(3/4,1/2)  A>0

 

AC – B^2  =2*3-1 = 6 – 1 = 5 > 0

(3/4,1/2) is local  minimum .

 

Similarly

At (1/3,-1/3)  we have  fxx=2   and  fxy=-1  and fyy=-2

AC-B^2

= 2 * -2  – (-1)^2 = -4  – 1  = 5<0

 

So saddle point  is (1/3,-1/3)

 

 

 

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