Trig Derivative - # 3

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Jason76

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\(\displaystyle y = \dfrac{9 - \sec x}{\tan x}\)

\(\displaystyle y' = \dfrac{[\tan x][-\sec x \tan x] - [9 - \sec x][sec^{2} x]}{\tan^{2} x}\) :confused:
 
Jason76 said:
\(\displaystyle y = \dfrac{9 - \sec x}{\tan x}\)

\(\displaystyle y' = \dfrac{[\tan x][-\sec x \tan x] - [9 - \sec x][sec^{2} x]}{\tan^{2} x}\) :confused:
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What is your question regarding this exercise? ;)
 
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