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

Jan 7, 2018

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

#### Explanation:

So, we first use the chain rule, which is:
$f ' \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$.
In this case, $f \left(x\right) = {\tan}^{-} 1 \left(x\right)$ and $g \left(x\right) = 3 {x}^{2} + 1$.
Given that $\frac{d}{\mathrm{dx}} {\tan}^{-} 1 \left(x\right) = \frac{1}{1 + {x}^{2}}$, we can continue.
1*(d/dx(3x^2+1))/(1+(3x^2+1)^2

Now we need to find out what $\frac{d}{\mathrm{dx}} g \left(x\right)$ is. Using the sum rule, which is d/dx (f(x)+g(x)) = d/dx f(x)+d/dx g(x)) where $f \left(x\right) = 3 {x}^{2}$ and $g \left(x\right) = 1$, we can continue.
$\frac{d}{\mathrm{dx}} g \left(x\right) = \frac{d}{\mathrm{dx}} 3 {x}^{2} + \frac{d}{\mathrm{dx}} 1$

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