find y and y. y = ln(|csc(x)|) y = y =
find y and y. y = ln(|csc(x)|) y = y =
Answer
Explanation:
Step1: Differentiate ( y = \ln(|\csc(x)|) ) using the chain rule
The derivative of ( \ln(u) ) with respect to ( x ) is ( \frac{u'}{u} ). Let ( u = |\csc(x)| ). The derivative of ( \csc(x) ) is ( -\csc(x)\cot(x) ). Since ( \frac{d}{dx}|\csc(x)|=\frac{\csc(x)(-\csc(x)\cot(x))}{|\csc(x)|} ) (using the formula ( \frac{d}{dx}|f(x)|=\frac{f(x)f'(x)}{|f(x)|} )), and for ( y = \ln(|\csc(x)|) ), ( y'=\frac{-\csc(x)\cot(x)}{\csc(x)}=-\cot(x) )
Step2: Differentiate ( y'=-\cot(x) ) to find ( y'' )
The derivative of ( \cot(x)=-\csc^{2}(x) ). So, using the constant - multiple rule ( (cf(x))' = cf'(x) ) where ( c=- 1 ), ( y''=\frac{d}{dx}(-\cot(x))=\csc^{2}(x) )
Answer:
( y'=-\cot(x) ) ( y''=\csc^{2}(x) )