final pratice exam help please no calculator

Question

Answer 1

12)

Given functions

g(3) = 2

g’(3)= 3

P(x) = [ g(x) ]²

Apply derivative both sides with respect to x.

P’(x) = 2g(x) [g’(x) ]

P’(x) = 2g(x)g’(x)

Substitute : x = 3

P’(3) = 2g(3)g’(3)

Substitute : g’(3)= 3 , g’(3)= 3

= 2(6)(3)

= 12

Solution : Option (d) is the Correct Answer.

Comments

when you say you apply derivate onto both side, which derivative formula did you use? The chain rule? Because you repeat the numbers twice. with the g(x)

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P’(x) = 2g(x)g’(x)

Substitute : x = 3

P’(3) = 2g(3)g’(3)

Substitute : g’(3)= 3 , g’(3)= 3 

= 2(6)(3)

= 12

 

looking at this again, I am confused now. IDK where the 6 come from. I actually very confused. Can you put more steps in between each?

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The Function is P(x) = [g(x)]².

Apply derivative on each sides with respect to x.

Now we make use of chain rule of derivative to differentiate the above function.

Chain rule of derivative is .

         (Power rule of Derivatve )

 

P'(x) = 2 g(x) * g'(x)

Substitute x = 3

P’(3) = 2 g(3) * g'(3)

Substitute g(3)= 2 , g’(3)= 3

P’(3) = 2*[2]*[3] = 4 * 3 = 12.

Therefore P'(3) = 12.

Answer 2

13)

Balloon volume filling rate V' = 4 ft³/min.

Radius r = 2 feet.

Balloon radius increasing rate r' = ?

Volume oh the Balloon is given by

Substitute : V' = 4 , r = 2.

Solution : Option (b) is the Correct Answer.

Comments

When you convert step 1 into step 2, why is there a r1? Where does that r1 come from? What derivative is that? and why V becomes v1 also?

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Here v' indicates the derivative of the volume with respect to t.

r' indicates the derivative of the radius with respect to t.

The volume of the balloon is .

Rate of change in the function is given by the derivative of that function.

Apply derivative on each sides with respect to t.

.

Answer 3

14)

Given equation is

x³ + 2xy + y² = 1

Apply derivative (implicit) both sides with respect to x.

3x² + 2 [ x(dy/dx) + y ] + 2y(dy/dx) = 0

3x² + 2x(dy/dx) + 2y + 2y(dy/dx) = 0

(2x + 2y) (dy/dx) = -3x² - 2y

dy/dx = - (3x² + 2y)/(2x + 2y)

Substitute : x = 1 , y = -2

dy/dx = - (3(1²) + 2(-2)) / (2(1) + 2(-2) )

dy/dx = - (3(1²) - 2(2)) / (2(1) - 2(2) )

dy/dx = - ( 3 - 4 ) / ( 2 - 4 )

dy/dx = - ( - 1 ) / ( - 2 )

dy/dx = - 1/2

Solution : Option (a) is the Correct Answer.

Comments

3x² + 2 [ x(dy/dx) + y ] + 2y(dy/dx) = 0

 

Why is there a dy/dx in between x and +y? Wasn't it xy before? how you just add a dy/dx inside them? Is that possible? 

 

Also what does dy/dx and what rule did you use?

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The function is x³ + 2xy + y² = 1.

To differentiate the above, we make use of the concept implicit differentiation.

Implicit differentiation means the process of finding the derivative of a dependent variable in an implicit function by differentiating each term separately.

As the term xy in the form of (uv), then to differentiate it we make use of product rule of differentiation.

Apply derivative on each sides with respect to x.

             (The derivative of a constant is zero)

   (Product rule of differentiation )

.

Answer 4

15)

Given function f '(-3) = 2

Apply anti-derivative (integration) with respect to x.

f (-3) = 2x + c

But f (-3) = 4

2x + c = 4

Substitute x = -3.

2(-3) + c = 4

-6 + c = 4

c = 4 + 6

c = 10

The function f (x ) = 2x + 10

Now Substitute x = -3.1.

f (-3.1) = 2(-3.1) + 10

f (-3.1) = - 6.2 + 10

f (-3.1) = 3.8

Solution : Option (a) is the Correct Answer.

Comments

"Apply anti-derivative (integration) with respect to x.

f (-3) = 2x + c

But f (-3) = 4

2x + c = 4"

 

I don't know how you started it. I don't know what is that anti-derivative thing. I also don't know how you get 2x + C.

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Alternative form:

The tangent line equation is in the form of y = mx + c

where y = f(x) and m = slope of the line is f'(x).

f(x) = (f'(x))x + c

Substitute x = -3

f(-3) = (f'(-3))*(-3)+ c

Given f(-3) = 4 and f'(-3) = 2

4 = (2)*(-3)+ c

4 = -6 + c

c = 6+4

c = 10

Therefore the tangent line equation is f(x) = 2x + 10.

f(-3.1) = 2*(-3.1) + 10

f(-3.1) = -6.2 + 10

f(-3.1) = 3.8