# Derivative of a Trigonometric Function Help

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 '14 at 0:08
The solution of the textbook is wrong actually. Please check the solution I posted. Good Luck! – Kushashwa Ravi Shrimali Jun 26 '14 at 1:04

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 '14 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 '14 at 1:12

Up to using the quotient rule is fine. You get something like
$$\newcommand{\cosec}{\operatorname{csc}}\\ \dfrac{dy}{dx} = \dfrac{-\cosec x\,\big(1+\cosec x\big)+\cot x\,\big(\cosec x \cot ⁡x\big)}{\big(1+\cosec x\big)^2 }$$

Using the identity $\,\cot^2x = \cosec^2 x-1$

\begin{align} \dfrac{dy}{dx} &= \dfrac{-\cosec x\,\big(1+\cosec x\big)+\cosec x\,\big(1-\cosec^2 x\big)}{\big(1+\cosec x\big)^2} \\ &= \dfrac{-\cosec x\,\big(1+\cosec x\big)+\cosec x\,\big(\big(1-\cosec x\big)\big(1+\cosec x\big)\big)}{\big(1+\cosec x\big)^2} \\ &= \dfrac{\big(1+\cosec x\big)\big(-\cosec x+\cosec x\,\big(1-\cosec x\big)\big)}{\big(1+\cosec x\big)^2} \\ &= \dfrac{-\cosec x+\cosec x-\cosec^2 x}{1+\cosec x} \\ &= \dfrac{-\cosec^2 x}{1+\cosec x} \end{align}

I got it in terms of $\,-\cosec^2 x$. Do check tour answer again.

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Please use MathJax to format your answer. It is almost incomprehensible otherwise. – Michael Albanese Dec 13 '15 at 6:07