Derivative of the functionhelp

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

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

That is,
[tex]\frac{dx}{dy}= -(1+ x^2)[/tex]
so
[tex]\frac{d cot^{-1}(x)}{dx}= \frac{dy}{dx}= \frac{1}{\frac{dx}{dy}}= \frac{-1}{x^2+ 1}[/tex].
 
ohh okay i get ur theory however
the answer still wrong...
its

-2x / 1 + (x^2 +2)^2
 
Last edited:
PMC_l0ver said:
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
[tex]\frac{-2x}{1} + (x^2 + 2)^2[/tex]

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)
 
Mark44 said:
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
[tex]\frac{-2x}{1} + (x^2 + 2)^2[/tex]

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
 
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 said:
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 said:
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:
[tex]dy/dx = \frac{-2x}{1 + (x^2 + 2)^2}[/tex]

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.