What is the second derivative of #f(x)=csc(3x^2-x)#?

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Answer 1 · 2018-03-03T03:00:38

#f'(x) = -csc(3x^2-x)cot(3x^2-x) * (6x-1)#

#f(x) = csc(3x^2-x)#
#f'(x) = -csc(3x^2-x)cot(3x^2-x) * d/dx(3x^2-x)# (cscx derivative and chain rule )

#f'(x) = -csc(3x^2-x)cot(3x^2-x) * (6x-1)#

derivative of #csc(x)#:

#csc(x) = 1/sin(x)#

#d/dx(csc(x)) = d/dx(1/sin(x))#

#=(sin(x)d/dx(1)-1(d/dx(sin(x))))/(sin^2(x))#

#=(sin(x)(0)-(cos(x)))/(sin^2(x))#

#=(-(cos(x)))/(sin^2(x))#

#=-1/sin(x) * cos(x)/sin(x)#

#=-csc(x)cot(x)#