Derivative of the functionhelp

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Homework Help Overview

The problem involves finding the derivative of the function y = cot-1(x2 + 2). Participants are discussing the correct application of derivative rules and the interpretation of the inverse cotangent function.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply derivative rules but questions their results against an answer key. Some participants clarify the distinction between cotangent and inverse cotangent, suggesting the need for a different approach to the derivative. Others express confusion over the formatting of the final answer and its interpretation in a web application.

Discussion Status

Participants are actively exploring different interpretations of the derivative and its representation. There is a focus on ensuring that the final answer is correctly formatted for submission. Some guidance has been offered regarding the use of parentheses in the expression.

Contextual Notes

There is an ongoing discussion about the terminology used for inverse functions and the implications for applying derivative rules. Participants are also considering the potential for misinterpretation in automated systems used for answer validation.

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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.
 

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