Some selected problems from 

Assignment 1 for chapters

5.8 - 5.10 &  6.1- 6.7

(First exam to be May 27th (4th week of class))


PS... If you have specific problems that  you want on here, type them up then e-mail them to me 

other wise I chose ones that I thought where of interest for some reason....  

most likely it was the algebra in the problem.  Any input to this site is appreciated!

Links that are helpful for doing the problems:

DIFFERENTIATION OF INVERSE TRIGONOMETRIC FUNCTIONS

Definition of Inverse Trigonometric Functions chart 

Inverses of Trigonometric Functions on Visual Basic- This shows how they are derived.

 


Working of problem side of  page

Thinking or Scratch side of page


Chapter 5 section 8 

Sec 5.8 pg 386 # 47

g(x) = (arcsin 3x)/ x   

First rearrange the problem 

to make it easier to work with.

g(x) = (1/x) * (arcsin 3x)

Things needed to do the problem.

(d/dx)[arcsin u] = [u' / (sqrt (1-u^2))]

let u = (3x) then u' = 3      

therefore:

(d/dx)[arcsin u] = [3 / (sqrt (1 - (3x)^2))]

 or  simplified:  [3 / (sqrt (1 - (9 x^2))]

Now diff of eqn:

g ' (x) = [ (d/dx)(1/x) *(arcsin 3x)] + [ (d/dx)(arcsin 3x) *(1/x)] 

g ' (x) = [ (-1/(x^2) *(arcsin 3x)] + [3 / (sqrt (1 - (9 x^2)) *(1/x)] 

g ' (x) = [ (-arcsin 3x)/(x^2) ] + [3 / (x*(sqrt (1 - (9 x^2)))] 

Rearrange to make it easier to work.

g ' (x) = [3 / (sqrt (1 - (9 x^2))] + [ (-arcsin 3x)/(x^2) ]

or  simplified further.

g ' (x) = [3 / (sqrt (1 - (9 x^2))] - [ (arcsin 3x)/(x^2) ]

Get a common denominator: 

[(x^2)*(sqrt (1 - (9x^2)))] 

Common denominator of 1st part: 

[3 / (x*(sqrt (1 - (9 x^2)))] * [x / x]

= [(3 * x )/(x ^2* (sqrt (1 - (9 x^2)))]

Common denominator of other part

[(arcsin 3x)/(x^2) ]* [(sqrt (1 - (9 x^2))) /

       (sqrt (1 - (9 x^2)))]

= [(sqrt (1 - (9 x^2)))*(arcsin 3x) / 

      (x^2 * (sqrt (1 - (9 x^2))))

Then combine and simplify

Final solution:

g ' (x) = [3 x - (sqrt (1 - (9x^2))) * (-arcsin 3x)]  / [(x^2)*(sqrt (1 - (9x^2)))


Sec 5.8 Page 386 # 53

y = (1/2)*[(1/2)Ln ((x+1)/(x-1)) + arctan x]

 

Things needed to do the problem.

(d/dx)[arctan u] = [u' / (1 + u^2)]

Let u = x,  then u ' = 1

therefore: 

(d/dx)[arctan u] = [1 / (1 + x^2)]

More things needed to do the problem.

(d/dx)[Ln x] = (1/x)

Simplify the problem, use rules about Natural Logs

Also factor out the constant in front of the Ln..

y = (1/2)*(1/2)*[(Ln(x+1) - Ln(x-1)] + [(1/2) arctan x]

Simplify the problem by combining (1/2)*(1/2).
y = (1/4)*[(Ln(x+1) - Ln(x-1)] + [(1/2) arctan x]

Now diff of eqn:

y ' = (1/4)*[1/(x+1) - 1/(x-1)] +  [(1/2) * (1 / (1 + x^2))]

Get a common denominator so terms can be combined  

(x^2 + 1)  

Common denominator 1st part

[(1/(x+1)) * (x-1)] = [(x-1) / (x^2-1)]

Common denominator 2nd part

[(1/(x-1)) * (x+1)] = [(x+1) / (x^2-1)]

 

Set back into equation.  

y ' = (1/4)*[((x-1)/(x^2-1)) - ((x+1)/(x^2-1))] +  [(1/2)*(1 /  (x^2 + 1))] 

Then combine and simplify

(1/4)*[-2/(x^2-1))] +  [(1/2)*(1 / (x^2 + 1))]

Simplify the problem more.

(-1/2)*[(1/(x^2+1))] +  [(1/2)*(1 / (x^2 + 1))]

Factor out (-1/2) to simplify

(-1/2)*[(1/(x^2-1)) +  (1 / (x^2 + 1))]

Get a common denominator so terms can be combined 

 (x^4 - 1)

Get a common denominator of 1st part

(1/(x^2-1)*[ (x^2+1)/(x^2+1)] 

= (x^2+1)/ (x^4 - 1)

Get a common denominator of 2nd part

(1 / (x^2 + 1))*[ (x^2-1)/(x^2-1)]

= (x^2-1)/(x^4 - 1)

(-1/2)*[((x^2+1)/ (x^4 - 1)) +  ((x^2-1)/(x^4 - 1))]

(-1/2)*[((x^2+1)*(x^2-1))/ (x^4 - 1)]

(-1/2)*[(-2)/ (x^4 - 1)] 

Final solution:

y ' (x) = [1/(x^4 - 1)] 


Sec 5.8 Page 386 # 57

y = 8 arcsin( x / 4) - [(x*(sqrt (16 - x^2)))/2]

Simplify the problem, change sqrt into exp of (1/2)

this makes it easier to use the chain rule.

y = [8 * arcsin( x / 4)] - [(x*(16 - x^2)^(1/2)))/2  ]

Things needed to do the problem.

(d/dx)[arcsin u] = [u' / (sqrt (1-u^2))]

Let u = ( x / 4),  then u ' = (1/4)

therefore: 

(d/dx)[arcsin u] = [(1/4) / (sqrt (1 - ( x / 4)^2))]

Now diff of eqn:

y ' (x) = [8 * (d/dx)(arcsin( x / 4))] - [((d/dx)x*(16 - x^2)^(1/2)))/2  ]+[(x*(d/dx)(16 - x^2)^(1/2)))/2  ]

y ' (x) = [8 *[(1/4) / (sqrt (1 - ( x / 4)^2))]] - [ (1)*((16 - x^2)^(1/2)))/2) + (x/2)*(1/2)*(-2x)*((16 - x^2)^(-1/2))]

Note, the derivative of 8 is zero therefore that part was not shown here since it would have cancelled out.

Now you need to simplify the problem as much as possible for the next few steps.

y ' (x) = [2* (1/ (sqrt (1 - ( x / 4)^2))] - [(sqrt (16 - x^2)))/2] - [(x/4)*(-2x)*((16 - x^2)^(-1/2)) ] 

y ' (x) = [2* (1/ (sqrt (1 - ( x ^2)/ 16)))] - [(sqrt (16 - x^2)))/2] + [(x^2) /((2)*(sqrt (16 - x^2))) ] 

y ' (x) = [2* (1/ (sqrt (16 - ( x ^2))/ 16)))] - [(sqrt (16 - x^2)))/2] + [(x^2) /((2)*(sqrt (16 - x^2))) ] 

y ' (x) = [4* (1/ (sqrt (16 - ( x ^2)))] - [(sqrt (16 - x^2)))/2] + [(x^2) /((2)*(sqrt (16 - x^2))) ] 

Now you need get a common denominator of (2 (sqrt (16 - ( x ^2))....  I will skip showing that step.

Final solution:

y ' (x) = ( x ^2) /(sqrt (16 - ( x ^2)


Page 387 # 73

This spot is reserved for problem 73

Coming soon!!!

 


Sec 5.8 Page 387 # 76

Verify each differentiation formula.

All of these are shown on the Visual Basic site:

Visual Basic- This shows how they are derived.

 

This is the last of the examples from Sec 5.8


 

Chapter 5 section 9

Sec 5.9 Page 393 # 25

Things needed to do the problem.

 ∫ (1 / ((x)^(1/2)*(1-x)^(1/2)))dx

 

You need to recognize the pattern form(s) of that equation.

Sometimes it help to move the dx to Numerator for ease of understanding the problem.
 ∫ (1*dx / ((x)^(1/2)*(1-x)^(1/2)))
∫ (dx / ((x)^(1/2)*(1-x)^(1/2)))

If we let u = (x)^(1/2) 

then du = 1/(2* (x)^(1/2))dx 

or 

dx = (2* (x)^(1/2))du

If x = u2 , then u =  ((u2)^(1/2)) 

this makes

dx = (2 * u)du

  The NEW form of the equation is like this:              

∫ ((2 * u)du / (u*(1-u2)^(1/2))

Cancel u in denominator and numerator

and move constant (2) out front of integral

This simplifies to :

 2* ∫ (du  /(1-u2)^(1/2))    

 

Now it  matches this form:

∫ (du / (a2 - u2 )^(1/2)= arcsin( u / a ) + C  

 2* ∫ (du /(1-u2)^(1/2))  

  = 2*arcsin( u / a ) + C 

Now re-substitute u and a. 

   = 2*arcsin( (x)^(1/2) / 1 ) + C 

This can be simplified for final answer.

Final solution:

  = 2*arcsin (x)^(1/2)  + C 


Sec 5.9 Page 393 # 29

Things needed to do the problem.

∫ ((x + 5)*dx / (9 - ( x -3)2)^(1/2)))

You need to recognize the pattern form(s) of that equation.

This equation  matches this form:

∫ (du / (a2 - u2 )^(1/2)= arcsin( u / a ) + C   

If you let a = 3 and u = ( x -3)

Since u = ( x - 3),  then du = 1 

But here would the numerator would be

(x + 5)* dx  so something more must be done.

We need ( x - 3) du  * dx 

Since (x + 5 ) - 8 = (x -3) we can use that.

 or      (x -3) + 8 = (x + 5 )

∫ ((x -3)*dx / (9 - ( x -3)2)^(1/2)))  +  ∫ (8dx / (9 - ( x -3)2)^(1/2)))

 

∫ (8 dx/ (9 - ( x -3)2)^(1/2)))  

You can find this integral (it fits the Arcsin Rule)

∫ (du / (a2 - u2 )^(1/2))= arcsin( u / a ) + C 

 or 8 arcsin (( x -3)/3) + C

 ∫ ((x -3)dx  / (9 - ( x -3)2)^(1/2)))

You can find this integral ( it fits the Power Rule)

 or   - (6x - x2 )^(1/2))

= - (6x - x2 )^(1/2)) + 8 arcsin (( x -3)/3) + C

 

8 arcsin (( x -3)/3) + C 

 can be simplified to

 8 arcsin (( x /3)+ 1) + C

Final solution:

- (6x - x2 )^(1/2)) + 8 arcsin (( x /3)+ 1) + C


Sec 5.9 Page 393 # 33

Things needed to do the problem.

∫ ((2x)dx / ( x 2 + 6x + 13))

If  u = ( x 2 + 6x + 13)

then du = (2x + 6)  

but we have (2x),

therefore we must add and 

subtract 6 in the numerator.

∫ ((2x + 6 - 6)dx / ( x 2 + 6x + 13))

This is not in a form where we can find the integral.

∫ ((2x + 6)dx / ( x 2 + 6x + 13)) + ∫ ((-6)* dx/ ( x 2 + 6x + 13))  

OR

∫ ((2x + 6)dx / ( x 2 + 6x + 13)) - 6 ∫ (dx/ ( x 2 + 6x + 13))

This is getting closer but the second equations

- 6 ∫ ((1)* dx/ ( x 2 + 6x + 13))  or

- 6 ∫ ((dx) / ( x 2 + 6x + 13))

 needs  some more adjustment to matches the form:

∫ ((du) / (a2   + u2  ))  = (1/a)* arctan (u/a) + C

This one will require that you complete the square.

( x 2 + 6x + 13) --> you need  6/2 = 3 then 32  = 9.

You must add 9 and subtract 9.

 = ( x 2 + 6x + 9) - 9 + 13

 = ( x 2 + 3)2  - 9 + 13

= ( x 2 + 3)2  + 4

= - 6 ∫ (dx / (( x 2 + 3)2  + 4))  or

= - 6 ∫ (dx / ( 4 + ( x 2 + 3)2  ))

In this case:  a = 4, u = ( x 2 + 3)

∫ (((2x + 6)*dx) / ( x 2 + 6x + 13) )) - 6∫ (dx / (4 + ( x  + 3)2  ))

Now we can get the integral for both parts of the whole equation.           

The 1st part of the equations

∫ (((2x + 6)*dx) / ( x 2 + 6x + 13) ))   

= Ln| x 2 + 6x + 13| 

The 2nd part of the equations

- 6∫ (((1)*dx) / (4 + ( x  + 3)2  ))  

= - 6arctan((x+3)/4) + C

Together the whole equation is

  = Ln| x 2 + 6x + 13| - 6arctan((x+3)/4) + C    

Final solution:

After simplification:

    = Ln| x 2 + 6x + 13| - 3arctan((x+3)/2) + C

 


This is the end of the problems for assignment in Section 5.9

If you want others added, e-mail the problems to me and I will post them.


Chapter 5 Section 10 

 I got too busy to type the problems so you will have to suffer with reading my writing.

  The good thing is that all assigned problems are shown instead of only a selected few.


Things needed to do the problem.


Things needed to do the problem.


Things needed to do the problem.

 

 


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