There are several I need help with.

Find the derivative

1.

2.

3.

Use the substitution rule

4.

5.

thanks

Printable View

- Jul 4th 2006, 01:40 PMc_323_hHelp w/ Problems for Test Tomorrow
There are several I need help with.

Find the derivative

1.

2.

3.

Use the substitution rule

4.

5.

thanks - Jul 4th 2006, 02:04 PMThePerfectHackerQuote:

Originally Posted by**c_323_h**

Let,

, thus

Thus,

Thus,

Substitute back,

- Jul 4th 2006, 02:07 PMThePerfectHackerQuote:

Originally Posted by**c_323_h**

then,

Thus,

Thus,

- Jul 4th 2006, 02:17 PMThePerfectHackerQuote:

Originally Posted by**c_323_h**

(Remember to use chain rule now)

Thus,

Thus,

Thus,

Thus,

Thus,

- Jul 4th 2006, 02:19 PMThePerfectHackerQuote:

Originally Posted by**c_323_h**

Now just use chain rule,

Product rule,

Express back,

- Jul 4th 2006, 02:23 PMThePerfectHackerQuote:

Originally Posted by**c_323_h**

Thus,

Thus,

- Jul 4th 2006, 02:33 PMc_323_h
Thank you very much for the quick response...and for solving all the problems.

- Jul 4th 2006, 02:59 PMSoroban
Hello, c_323_h!

Quote:

Find the derivative:

Differentiate implicitly: .

Then we have: .

Rearrange terms: .

Factor: .

Therefore: .

Quote:

Quote:

Take logs: .

Differentiate implicitly: .

We have: .

Since , we have: .

Quote:

Use the substitution rule:

We have: .

Let:

Substitute: .

You can finish it now . . .

Quote:

We have: .

Let:

Substitute: . . . . got it?

- Jul 4th 2006, 03:02 PMc_323_h
thanks you. as you can tell i am struggling with this class. is anyone willing to help me with these?

Differentiate:

1.

2.

Evaluate the integral:

1. - Jul 4th 2006, 07:04 PMThePerfectHackerQuote:

Originally Posted by**c_323_h**

As, (since ) (I factored it)

Let, then, .

Thus, far we have,

Since, then, thus, thus, Thus, we have,

This is a inverse hyperbolic function which you should recognize thus,

Substitute back,

Thus,

- Jul 4th 2006, 07:19 PMc_323_hQuote:

Originally Posted by**ThePerfectHacker**

- Jul 4th 2006, 09:24 PMSoroban
Hello again, c_323_h!

Quote:

Differentiate:

Quote:

. . . . .******

Quote:

Evaluate the integral: .

In the denominator, complete the square:

The integral becomes: .

. . Let

. . Note that: .

. . . . . . . . . .

Substitute: .

Back-substitute

We have: .

So we have: .

. . .

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

******

In #2, we have: .

Let: . . . and:

Then . . . and

Since , is in this right triangle:Code:`*`

/β|

/ |1

/α |

* - - - *

t

Since is the*other acute angle.*

That is: .

Hence, the function is: .

No wonder the derivative is 0 . . .