Calculus Help

[BullyCanadian]

Here is the question. I am gonna go get some help tomorrow, but was wondering if anyone can help me out right now (I know most of you guys are really smart)

Here is question 1
The question is to show that the equation on the left equals the right

EDIT - There is a mistake in the question it should be cos{arctan[sin(arcot x)]} = ....

 

[KLin]

x = 42

 

[Gibson486]

well, what do you want us to do with the problem????

 

[TechnoKid]

Originally posted by: KLin
x = 42

really?

 

[BullyCanadian]

Originally posted by: Gibson486
well, what do you want us to do with the problem????

Ohh sorry about that,

Basically Show that, the first equals the second.

 

[Semidevil]

so are we supposed to differentiate something or integrate something? what calculus is this?

 

[BullyCanadian]

Originally posted by: Semidevil
so are we supposed to differentiate something or integrate something? what calculus is this?

calc 102 first year science in uni.

yes I believe to differentiate, but i have idea about where to start :S trig makes me scared. I have a tutor ready for tomorrow, but need to get started on this question today.

 

[BullyCanadian]

up

 

[Schrodinger]

UofT or York?

 

[BullyCanadian]

Mac

 

[Gibson486]

http://www.math.com/tables/derivatives/tableof.htm


look at the table for inverse derivatives for now....


give me a sec so i can find a better site to explain it.

 

[chuckywang]

Originally posted by: BullyCanadian
Here is the question. I am gonna go get some help tomorrow, but was wondering if anyone can help me out right now (I know most of you guys are really smart)

Here is question 1
The question is to show that the equation on the left equals the right

But the left does not equal the right.

Let x = 1, and you'll see the two sides don't equate.

 

[Gibson486]

http://www.sosmath.com/calculus/diff/der08/der08.html

here is a better explanation.

 

[Gibson486]

Originally posted by: chuckywang

Originally posted by: BullyCanadian
Here is the question. I am gonna go get some help tomorrow, but was wondering if anyone can help me out right now (I know most of you guys are really smart)

Here is question 1
The question is to show that the equation on the left equals the right

But the left does not equal the right.

Let x = 1, and you'll see the two sides don't equate.

indeed, 1 != root(2/3)

 

[BullyCanadian]

Originally posted by: Gibson486

Originally posted by: chuckywang

Originally posted by: BullyCanadian
Here is the question. I am gonna go get some help tomorrow, but was wondering if anyone can help me out right now (I know most of you guys are really smart)

Here is question 1
The question is to show that the equation on the left equals the right

But the left does not equal the right.

Let x = 1, and you'll see the two sides don't equate.

indeed, 1 != root(2/3)

My mistake! I wrote it wrong! missed sin before arccot

 

[chuckywang]

Originally posted by: BullyCanadian

EDIT - There is a mistake in the question it should be cos{arctan[sin(arcot x)]} = ....

Do this problem in two parts:

First, compute y=sin(arccot x) = 1/sqrt(x^2 + 1)

Then, compute cos(arctan y) = 1/sqrt(y^2 + 1)

Plug in y= 1/sqrt(x^2 + 1) to 1/sqrt(y^2 + 1)
and you get 1/sqrt(1/(x^2+1) + 1) = sqrt(x^2 + 1)/sqrt(x^2 + 2).

 

[chuckywang]

Originally posted by: Gibson486

Originally posted by: chuckywang

Originally posted by: BullyCanadian
Here is the question. I am gonna go get some help tomorrow, but was wondering if anyone can help me out right now (I know most of you guys are really smart)

Here is question 1
The question is to show that the equation on the left equals the right

But the left does not equal the right.

Let x = 1, and you'll see the two sides don't equate.

indeed, 1 != root(2/3)

More like 1/sqrt( (pi/4)^2 + 1) != sqrt(2/3)

 

[BullyCanadian]

Originally posted by: chuckywang

Originally posted by: BullyCanadian

EDIT - There is a mistake in the question it should be cos{arctan[sin(arcot x)]} = ....

Do this problem in two parts:

First, compute y=sin(arccot x) = 1/sqrt(x^2 + 1)

Then, compute cos(arctan y) = 1/sqrt(y^2 + 1)

Plug in y= 1/sqrt(x^2 + 1) to 1/sqrt(y^2 + 1)
and you get 1/sqrt(1/(x^2+1) + 1) = sqrt(x^2 + 1)/sqrt(x^2 + 2).

ok I get everything up to this:

and you get 1/sqrt(1/(x^2+1) + 1) = sqrt(x^2 + 1)/sqrt(x^2 + 2)

how does /sqrt(1/(x^2+1) + 1) = sqrt(x^2 + 1)/sqrt(x^2 + 2)???

 

[chuckywang]

Originally posted by: BullyCanadian

Originally posted by: chuckywang

Originally posted by: BullyCanadian

EDIT - There is a mistake in the question it should be cos{arctan[sin(arcot x)]} = ....

Do this problem in two parts:

First, compute y=sin(arccot x) = 1/sqrt(x^2 + 1)

Then, compute cos(arctan y) = 1/sqrt(y^2 + 1)

Plug in y= 1/sqrt(x^2 + 1) to 1/sqrt(y^2 + 1)
and you get 1/sqrt(1/(x^2+1) + 1) = sqrt(x^2 + 1)/sqrt(x^2 + 2).

ok I get everything up to this:

and you get 1/sqrt(1/(x^2+1) + 1) = sqrt(x^2 + 1)/sqrt(x^2 + 2)

how does /sqrt(1/(x^2+1) + 1) = sqrt(x^2 + 1)/sqrt(x^2 + 2)???

1/sqrt(1/(x^2+1) + 1) = sqrt(1/(1/(x^2+1)+1) = sqrt( (x^2+1)/(1+x^2+1)) = sqrt((x^2+1)/(x^2+2)

 

[BullyCanadian]

thank you very much!