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 $csc(x)$ :

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

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

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

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

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

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

$=-csc(x)cot(x)$

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