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