Question

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

Answer 1

`(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`