What is the second derivative of $f (x) = cot (3 x^{2} - x)$ $f(x)=cot(3x^2-x)$ ?

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Answer 1 · 2015-12-30T17:34:49

First derivative:

Use the chain rule, which states that $\frac{d}{d x} (cot (u)) = - {csc}^{2} (u) \cdot \frac{d u}{d x}$ $d/dx(cot(u))=-csc^2(u)*(du)/dx$

Thus,

$f'(x)=-csc^2(3x^2-x)*d/dx(3x^2-x)$

$=>-(6x-1)csc^2(3x^2-x)$

Second derivative:

Use product rule in conjunction with the chain rule again.

When doing chain rule with the cosecant function squared, the overriding issue will be the exponent, and then the cosecant.

$f''(x)=-csc^2(3x^2-x)d/dx(6x-1)-(6x-1)d/dx(csc^2(3x^2-x))$

Find each derivative.

$d/dx(6x-1)=6$

$d/dx(csc^2(3x^2-x))=2csc(3x^2-x)d/dx(csc(3x^2-x))$

$color(white)(ssss)$ Recall that $d/dx(csc(u))=-csc(u)cot(u)*(du)/dx$ .

$=>2csc(3x^2-x) * -csc(3x^2-x)cot(3x^2-x) * (6x-1)$

$=>-2(6x-1)csc^2(3x^2-x)cot(3x^2-x)$

Plug both the derivatives back in to find $f''(x)$ .

$f''(x)=-6csc^2(3x^2-x)+2(6x-1)^2csc^2(3x^2-x)cot(3x^2-x)$

This can be further simplified, if you want:

$f''(x)=((72x^2-24x+2)cot(3x^2-x)-6)csc^2(3x^2-x)$

What is the second derivative of f(x)=cot(3x^2-x)f(x)=cot(3x2−x)f(x)=cot(3x^2-x)?