Trig Derivative - # 2

• Oct 14th 2013, 12:05 PM
Jason76
Trig Derivative - # 2
\$\displaystyle y = \dfrac{8x}{1 - \cot x}\$

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

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

\$\displaystyle y' = \dfrac{[8(1 - \cot x)]-[-8 \csc^{3} x]}{(1 - \cot x)^{2}}\$ http://www.freemathhelp.com/forum/im...s/confused.png
• Oct 14th 2013, 12:29 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 \$\displaystyle 1-\cot x\$. Then \$\displaystyle (8x)(\csc^2 x) \neq 8\csc^3 x\$
• Oct 14th 2013, 02:09 PM
Jason76
Re: Trig Derivative - # 2
Ok, let's look at his again:

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

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

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

\$\displaystyle \dfrac{8 - 8\cot x - 8x - \csc^{2} x}{(1 - \cot x)^{2}}\$ - This answer is still wrong on computer homework. (Speechless)
• Oct 14th 2013, 03:36 PM
SlipEternal
Re: Trig Derivative - # 2
What is the derivative of \$\displaystyle 1-\cot x\$? It is not \$\displaystyle 1-(-\csc^2 x)\$. Even if it were, \$\displaystyle (8x)(1+\csc^2 x) \neq 8x + \csc^2x\$.
• Oct 15th 2013, 11:59 AM
Jason76
Re: Trig Derivative - # 2
I figured out the right answer. The derivative of \$\displaystyle 1 - \cot x = -(-csc x)\$ or \$\displaystyle \csc x\$ So you just put that in there, and you get the right answer.