differentiate

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March 9th 2009, 11:37 AM
algebra2

differentiate

how do you differentiate ln|cscx + cotx|
March 9th 2009, 11:54 AM
Soroban

Hello, algebra2!

Quote:

How do you differentiate: . $f(x) \:=\:\ln|\csc x + \cot x|$

Chain Rule: . $f'(x) \;=\;\frac{1}{\csc x + \cot x}\cdot(\text{-}\csc x\cot x - \csc^2\!x)$

Factor: . $f'(x) \;=\;\frac{\text{-}\csc x\left(\cot x + \csc x\right)}{\csc x + \cot x}$

Reduce: . $f'(x) \;=\;\text{-}\csc x$
March 9th 2009, 01:02 PM
Air

Quote:

Originally Posted by algebra2

how do you differentiate ln|cscx + cotx|

Also note that $\int \ \frac{f'(x)}{f(x)}= \ln|f(x)| + c$ . For example $\int \ \frac{1}{x}= \ln|x| + c$ thus $\frac{\mathrm{d}\left(\ln |x|\right)}{\mathrm{d}x} = \frac{1}{x}$

Hence: $\frac{\mathrm{d}\left(\ln|\csc x + \cot x |\right)}{\mathrm{d}x} = \frac{\left(\csc x + \cot x \right)'}{\csc x + \cot x} = ...$

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