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Trying to derive a trigonometric function, Wolfram Alpha and my textbook provide two different answers. Here is the function:

$$y = {\cot x\over (1+\csc x)}$$

First step using quotient rule results in :

$${dy\over dx}={-\csc^2x(1+\csc x) - \cot x(-\csc x \cot x)\over (1+\csc x)^2}$$

from there I broke it into a difference of derivatives:

$$={-\csc x(1+\csc x)\over (1+\csc x)^2}+{\cot x(\csc x \cot x)\over (1+\csc x)^2}$$

The solution provided by my text book says the solution is what the first rational expression in the previous difference simplifies to:

$$= {-\csc x\over 1+\csc x}$$

So, am I just missing how $\dfrac{\cot x(\csc x \cot x)}{ (1+\csc x)^2}$ goes to null?

Thanks for the help!

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Either you have an error in the original function or your text book is wrong. The integral of the text book solution is not equal to the original function. –  Valtteri Jun 26 at 0:08
    
The solution of the textbook is wrong actually. Please check the solution I posted. Good Luck! –  Kushashwa Ravi Shrimali Jun 26 at 1:04

1 Answer 1

up vote 1 down vote accepted

The quotient rule states that : $$ \cfrac{d }{dx}\left(\cfrac{u}{v}\right) = \cfrac{ u'v - uv'}{v^2} $$ So, by using quotient rule in $y = \cfrac{\cot x}{ ( 1+\csc x)^2} $ gives : $$\cfrac{-\csc^2 x \left( 1 + \csc x \right)^2 - \cot x \left( 2 \left(1 + \csc x\right) \times \left(- \csc x \cot x \right)\right)}{(1 +\csc x)^4}\\$$ Take (1 + csc x) common

$$\implies \cfrac{\left(1 + \csc x\right) ( -\csc^2 x (1+\csc x) + 2\cot^2 x \csc x}{(1+\csc x)^4} \\ \implies \cfrac{(-\csc^2 x (1+\csc x) + 2\cot^2 x \csc x}{(1+\csc x)^3} $$

Now take $ -\csc x$ common, you get :

$$\cfrac{-\csc x(\csc x + csc^2 x - 2\cot^2 x}{(1+\csc x)^3} \\ \cfrac{-\csc x(\csc x + \color{blue}{\left(1+ \cot^2 x\right)} - 2\cot^2 x}{(1+\csc x)^3} \\ \text{Since : } \space \boxed{\csc^2 x = 1+ \cot^2 x} \\ \cfrac{-\csc x( \csc x - \cot^2 x + 1)}{(1+\csc x)^3} $$

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thank you! Although I'm not sure why my comments aren't showing up, but I mis-typed the problem. The denominator of the original function is not squared (I have edited the post to show so). I tried to enter an answer, but since I'm new it would not allow me to. –  PathemaMike Jun 26 at 1:08
    
Okay, no problems. You can mark this as answer by clicking the checkmark so that others may know well about this. Good Luck! –  Kushashwa Ravi Shrimali Jun 26 at 1:12

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