Derivative of the functionhelp

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In summary, a derivative of a function is a measure of the rate of change of the function with respect to the input. It is important because it helps us understand the behavior of the function and can be calculated using the limit definition or various derivative rules. Real-life applications of the derivative include physics, economics, and engineering, and it can be used to solve problems by finding maximum or minimum values, determining rates of change, and analyzing function behavior.
  • #1
PMC_l0ver
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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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  • #2
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].
 
  • #3
ohh okay i get ur theory however
the answer still wrong...
its

-2x / 1 + (x^2 +2)^2
 
Last edited:
  • #4
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)
 
  • #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
 
  • #6
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 >.<
 
  • #7
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.
 

FAQ: Derivative of the functionhelp

What is a derivative of a function?

A derivative of a function is a measure of the rate at which the output of the function changes with respect to the input. It is essentially the slope of the function at a specific point.

Why is the derivative of a function important?

The derivative of a function is important because it helps us understand the behavior of the function. It can tell us the direction of the function's change, whether it is increasing or decreasing, and the rate at which it is changing.

How is the derivative of a function calculated?

The derivative of a function can be calculated using the limit definition of a derivative or through various derivative rules and formulas, such as the power rule, product rule, and chain rule.

What are some real-life applications of the derivative of a function?

The derivative of a function has many real-life applications, such as in physics (calculating velocity and acceleration), economics (maximizing profits and minimizing costs), and engineering (determining optimal designs and rates of change).

How can I use the derivative of a function to solve problems?

The derivative of a function can be used to solve problems by helping us find the maximum or minimum value of a function, determining the rate of change of a quantity, and analyzing the behavior of a function at a specific point.

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