SOLUTION: Find dy/dx. Then use trigonometry to rite the derivative as a single trig function.
A) y = ln |cscx - cotx|
B) y = ln |sinx|
C) y = ln |secx + tanx|
D) y = -ln |cosx|
E) Pr
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Question 1030346: Find dy/dx. Then use trigonometry to rite the derivative as a single trig function.
A) y = ln |cscx - cotx|
B) y = ln |sinx|
C) y = ln |secx + tanx|
D) y = -ln |cosx|
E) Prove that -ln |cosx| = ln |secx|
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
I'll do parts A) and B) to get you started. I'll leave parts C) through E) for you to attempt on your own. If you get stuck, then you can post a separate question.
----------------------------------------------------------------------------------------------------------------------
Part A)
 - \cot(x)|)
 - \cot(x)}*\frac{d}{dx}\left(\csc(x) - \cot(x)\right))
 - \cot(x)}*\left(-\csc(x)*\cot(x) + \csc^2(x)\right))
 - \cot(x)}*\csc(x)\left(\csc(x)-\cot(x)\right))
*\frac{\csc(x)-\cot(x)}{\csc(x) - \cot(x)})
)
----------------------------------------------------------------------------------------------------------------------
Part B)
|)
}*\frac{d}{dx}\left(\sin(x)\right))
}*\cos(x))
}{\sin(x)})
)
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