How do you find the derivative of #tan^-1 (3x^2 +1)#?

1 Answer
Jan 7, 2018

#(6x)/(1+(3x^2+1)^2#

Explanation:

So, we first use the chain rule, which is:
#f'(g(x)) = f'(g(x)) * g'(x)#.
In this case, #f(x) = tan^-1(x)# and #g(x)=3x^2+1#.
Given that #d/dx tan^-1(x) = 1/(1+x^2)#, we can continue.
#1*(d/dx(3x^2+1))/(1+(3x^2+1)^2#

Now we need to find out what #d/dx g(x)# is. Using the sum rule, which is #d/dx (f(x)+g(x)) = d/dx f(x)+d/dx g(x))# where #f(x)=3x^2# and #g(x)=1#, we can continue.
#d/dx g(x)= d/dx 3x^2 + d/dx 1#

I'm going to skip the step where you derive what #d/dx 3x^2# is, but the answer to that is #6x#.
#d/dx g(x)= 6x + 0#
#d/dx g(x) = 6x#
Plugging that in, we get:
#1*(6x)/(1+(3x^2+1)^2#
#(6x)/(1+(3x^2+1)^2#