Derivative of the functionhelp

[PMC_l0ver]

Homework Statement

find dy/dx in as simplified a form as possible

y = Cot^-1 (x² + 2)

Homework Equations

identities
derivatives

The Attempt at a Solution

dy/dx = -1 (cot^-2(x²+2)(-csc²(x²+2)2x

dy/dx = (-1 (-csc²(x²+2)2x) / cot²(x²+2)

i also did the quotient rule for derivative and it gives me the same answer
however when i look at the answer key my answer is wrong...
am i missing something?

it seems like csc and cot must cancel out in this problem
please help me

 

[HallsofIvy]

The problem is that cot^{-1}(x)] is NOT cotangent to the -1 power and so the power rule is not applicable here. cot^{-1}(x) is the "inverse cotangent" (yes, it's an annoying conflict with terminology).

That is, if y= cot^{-1}(x) then x= cot(y). The derivative of cot(y) is "-csc^2(y)". Now you can use a trig identity, 1+ cot^2(y)= csc^1(y), to write that as -(1+ cot^2(y))= -(1+ x^2).

That is,
\frac{dx}{dy}= -(1+ x^2)
so
\frac{d cot^{-1}(x)}{dx}= \frac{dy}{dx}= \frac{1}{\frac{dx}{dy}}= \frac{-1}{x^2+ 1}.

 

[PMC_l0ver]

ohh okay i get ur theory however
the answer still wrong...
its

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

 

[Mark44]

ohh okay i get ur theory however
the answer still wrong...
its

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

Are you entering your answer into a Web application that compares what you enter with the correct answer? If so, the above is incorrect. It would be interpreted as
\frac{-2x}{1} + (x^2 + 2)^2

It should be written with parentheses around the entire denominator, like this:
-2x / (1 + (x^2 +2)^2)

or better yet, with the denominator expanded, like this:
-2x/(x^4 + 4x^2 + 5)

 

[PMC_l0ver]

Are you entering your answer into a Web application that compares what you enter with the correct answer? If so, the above is incorrect. It would be interpreted as
\frac{-2x}{1} + (x^2 + 2)^2

It should be written with parentheses around the entire denominator, like this:
-2x / (1 + (x^2 +2)^2)

or better yet, with the denominator expanded, like this:
-2x/(x^4 + 4x^2 + 5)

nope that i posted was right
its based on my txt book
please help me

 

[PMC_l0ver]

here what it says on the answer key

dy/dx = ( -1 / 1+(x^2+2)^2) 2x

dy/dx= -2x / 1+(x^2+2)^2


i've been trying to figure this problem for about 4hours already >.<

 

[Mark44]

It should be written with parentheses around the entire denominator, like this:
-2x / (1 + (x^2 +2)^2)
or better yet, with the denominator expanded, like this:
-2x/(x^4 + 4x^2 + 5)

here what it says on the answer key

dy/dx = ( -1 / 1+(x^2+2)^2) 2x

dy/dx= -2x / 1+(x^2+2)^2

Compare what I wrote (copied above) with what you are saying is in the answer key. The second form you wrote probably appears like this:
dy/dx = \frac{-2x}{1 + (x^2 + 2)^2}

If you are writing this in text (and not using LaTeX), you need parentheses around the entire denominator, like this:
dy/dx= -2x / (1+(x^2+2)^2)

This is the same as what I wrote.