Question

Hey all,
I am working on #3g of 5A Problem Set 2 and I am having trouble understanding how they determined the derivative of the inverse trig function arctan(x/sqrt(1-x^2)). By definition, d/dx(arctan(x))=1/(1+x^2). Also, from the soln., I see that they took all the junk in the parentheses and called all of that "y," but isn't this function y=arctan(x/(sqrt(1-x^2))) to begin with, making it also tan(y)=x/(sqrt(1-x^2))?

Answer 1

First, you ask,
*** isn't this function y=arctan(x/(sqrt(1-x^2))) to begin with? ***

Short answer: no.
Longer answer: it is a convention to write y= f(x), and so you might think
y = arctan( stuff)
But in this case, they are using "y" as just another variable with no special meaning other than to say
\[ y= \frac{x}{\sqrt{1+x^2}}\]

which they do so they can make the write-up easier to read.

To avoid this confusion, we can use "u" instead. Let
\[ u= \frac{x}{\sqrt{1+x^2}}\]
By the "chain rule"
\[ \frac{d}{dx} \tan^{-1}u = \frac{d}{du} \tan^{-1}u \cdot \frac{d}{dx} u \]

Answer 2

We get
\[ \frac{d}{dx} \tan^{-1}u = \frac{d}{du} \tan^{-1}u \cdot \frac{d}{dx} u \\
= \frac{1}{1+u^2} \cdot \frac{d}{dx} \left( \frac{x}{\sqrt{1+x^2}}\right)
\]

we can replace u^2 with its definition and simplify the 1/(1+u^2)
we can take the derivative of the other term.
simplify to get the answer.


Notice that the chain rule is needed to do this problem.... and it gets a bit messy

Answer 3

Okay, that makes a whole lot of sense. Thank you for your reply and sorry it took me a bit to reply:]