# Trig Derivative - # 2

• October 14th 2013, 01:05 PM
Jason76
Trig Derivative - # 2
$y = \dfrac{8x}{1 - \cot x}$

$y' = \dfrac{[1 - \cot x][8]-[8x][-\csc^{2} x]}{(1 - \cot x)^{2}}$

$y' = \dfrac{[8 - 8 \cot x]-[-8 \csc^{3} x]}{(1 - \cot x)^{2}}$

$y' = \dfrac{[8(1 - \cot x)]-[-8 \csc^{3} x]}{(1 - \cot x)^{2}}$ http://www.freemathhelp.com/forum/im...s/confused.png
• October 14th 2013, 01:29 PM
SlipEternal
Re: Trig Derivative - # 2
Not quite. You made a mistake on line 2, then another one on line 3. Recalculate the derivative of $1-\cot x$. Then $(8x)(\csc^2 x) \neq 8\csc^3 x$
• October 14th 2013, 03:09 PM
Jason76
Re: Trig Derivative - # 2
Ok, let's look at his again:

$y = \dfrac{8x}{1 - \cot x}$

$\dfrac{[1 - \cot x][8] - [8x][1 - (- \csc^{2} x)]}{(1 - \cot x)^{2}}$

$\dfrac{[1 - \cot x][8] - [8x][1 + \csc^{2} x]}{(1 - \cot x)^{2}}$

$\dfrac{8 - 8\cot x - 8x - \csc^{2} x}{(1 - \cot x)^{2}}$ - This answer is still wrong on computer homework. (Speechless)
• October 14th 2013, 04:36 PM
SlipEternal
Re: Trig Derivative - # 2
What is the derivative of $1-\cot x$? It is not $1-(-\csc^2 x)$. Even if it were, $(8x)(1+\csc^2 x) \neq 8x + \csc^2x$.
• October 15th 2013, 12:59 PM
Jason76
Re: Trig Derivative - # 2
I figured out the right answer. The derivative of $1 - \cot x = -(-csc x)$ or $\csc x$ So you just put that in there, and you get the right answer.