Thread: Derivative of Trigonometric Function

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

$

Find $\displaystyle f'(x)$

Originally Posted by StarlitxSunshine

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

$

Find $\displaystyle f'(x)$

You know how to find the derivatives of $\displaystyle cot(x)$ and $\displaystyle csc(x)$ 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:

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

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

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

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

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

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

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 >_<

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.

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:

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