# Thread: Trig Derivative - # 2

1. ## Trig Derivative - # 2

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

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

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

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

2. ## 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 $(8x)(\csc^2 x) \neq 8\csc^3 x$

3. ## 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.

4. ## Re: Trig Derivative - # 2

What is the derivative of $1-\cot x$? It is not $1-(-\csc^2 x)$. Even if it were, $(8x)(1+\csc^2 x) \neq 8x + \csc^2x$.

5. ## 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.