How do you find the derivative of the function: #(arctan(6x^2 +5))^2#?

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Answer 1 · 2016-01-23T15:59:31

#(24xarctan(6x^2+5))/((6x^2+5)^2+1)#

This will require a little chain rule.

The first issue is the second power, which can be dealt with as such: #d/dx[u^2]=2u*u'#. We know that #u=arctan(6x^2+5)#, so

#f'(x)=2arctan(6x^2+5)d/dx[arctan(6x^2+5)]#

Now, we must deal with the #arctan# function, which uses the rule #d/dx[arctan(u)]=(u')/(u^2+1)#, and we have #u=6x^2+5#, giving us

#f'(x)=2arctan(6x^2+5)*(d/dx[6x^2+5])/((6x^2+5)^2+1)#

Simplify:

#f'(x)=(2arctan(6x^2+5)*12x)/((6x^2+5)^2+1)#

#f'(x)=(24xarctan(6x^2+5))/((6x^2+5)^2+1)#