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

Question

Emily Ammerman · Accepted Answer

First derivative:

Use the chain rule, which states that #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)#

Emilia Aguilar · Answer

undefined To find the second derivative of $ f(x) = \cot(3x^2 - x) $, first, find the first derivative, then differentiate it again.

First derivative:
$$ f'(x) = -\csc^2(3x^2 - x)(6x - 1) $$

Second derivative:
$$ f''(x) = -\frac{d}{dx}(\csc^2(3x^2 - x)(6x - 1)) $$
$$ = -\frac{d}{dx}(-\csc^2(3x^2 - x)(6x - 1)) $$
$$ = -\frac{d}{dx}(-6x\csc^2(3x^2 - x) + \csc^2(3x^2 - x)) $$
$$ = 6\csc^2(3x^2 - x) - 12x\csc(3x^2 - x)\cot(3x^2 - x) - 2\csc^4(3x^2 - x)(6x - 1) $$

So, the second derivative of $ f(x) = \cot(3x^2 - x) $ is:
$$ f''(x) = 6\csc^2(3x^2 - x) - 12x\csc(3x^2 - x)\cot(3x^2 - x) - 2\csc^4(3x^2 - x)(6x - 1) $$