Derivative of Trigonometric Function

**StarlitxSunshine** · Nov 5th 2009, 03:59 PM

$ f(x) = \frac{cot x}{(1 + csc x)} $

Find $f'(x)$

**adkinsjr** · Nov 5th 2009, 05:00 PM

Originally Posted by StarlitxSunshine

$ f(x) = \frac{cot x}{(1 + csc x)} $

Find $f'(x)$

You know how to find the derivatives of $cot(x)$ and $csc(x)$ don't you? It's pretty easy if you know the derivatives of sine and cosine, so I won't go through it, I'll just write them down:

$\frac{d}{dx}cot(x)=-csc^2(x)$

$\frac{d}{dx}csc(x)=-csc(x)cot(x)$

So to find $f'(x)$ , just apply the quotient rule.

$f'(x)=\frac{[1+csc(x)][-csc^2(x)]-cot(x)[-csc(x)cot(x)]}{[1+csc(x)]^2}$

$=\frac{[1-csc^2(x)]+cot^2(x)csc(x)}{[1+csc(x)]^2}$

EDIT: The last line is WRONG. I didn't use the distributive property on the first term. When I mulitplied $1+csc(x)$ by $-csc^2(x)$ I should have obtained $-csc^2(x)-csc^3(x)$ .

**StarlitxSunshine** · Nov 5th 2009, 05:04 PM

Yup Yup :3 I know those derivatives.

The thing is, I got the first step (using the quotient rule), but after that, I'm not sure how you simplified to get the answer ?

Sorry for the trouble of explaining it twice >_<

**adkinsjr** · Nov 5th 2009, 05:07 PM

Originally Posted by StarlitxSunshine

Yup Yup :3 I know those derivatives.

The thing is, I got the first step (using the quotient rule), but after that, I'm not sure how you simplified to get the answer ?

Sorry for the trouble of explaining it twice >_<

In the numerator

lol, the reason why is I simplified it wrong. Sorry about that. I'll fix it in a minute.

**StarlitxSunshine** · Nov 5th 2009, 05:16 PM

Ohh... I think I understand what you did :33

But I have the answer to the question && it doesn't match up.

The problem is that I can't get from the quotient rule step to the answer.

Sorry for complicating it so much >_<

The Answer:

$\frac{-csc x}{1+cscx}$

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