What is the derivative of $f(x) = arctan(1 + x^3)$ ?

Question

1569 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-20T23:43:07

We know by definition that when $y=arctan(f)$ , we have its derivative as

$y'=(f')/(1+f²)$

Using the chain rule, which states that

$(dy)/(dx)=(dy)/(du)*(du)/(dx)$

we can rename $u=1+x^3$ and then start working with $f(x)=arctan(u)$ instead.

Then, respecting the chain rule:

$(dy)/(du)=(u')/(1+u^2)$

$(du)/(dx)=3x^2$

Now,

$(dy)/(dx)=((u')/(1+u^2))3x^2$

Let's substitute $u$

$(dy)/(dx)=((3x^2)/(1+(1+x^3)^2))3x^2=(9x^4)/(1+1+2x^3+x^6)=color(green)((9x^4)/(x^6+2x^3+2))$

What is the derivative of f(x) = arctan(1 + x^3)f(x) = arctan(1 + x^3)?