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

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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))#