# Thread: product of inverse trig function

1. ## product of inverse trig function

I need help with taking the derivative with $y=\frac{arctanx}{x}$. I cannot seem to even begin. In the solution manual it does the product derivative which I try doing, but I cannot figure out how to do it with the inverse of the tan function. I don't know what to do and the solution manual just jumps to the last step before the solution to "show" its work. Any hints?

2. ## Re: product of inverse trig function

Originally Posted by Barthayn
I need help with taking the derivative with $y=\frac{arctanx}{x}$.
$y=\frac{\arctan(x)}{x}=(x^{-1})\arctan(x)}$ use product rule.

$\frac{d[\arctan(x)]}{dx}=\frac{1}{1+x^2}$

3. ## Re: product of inverse trig function

I understand I have to use the product rule, but I don't see how $\frac{d[\arctan(x)]}{dx}=\frac{1}{1+x^2}$.

I get ${tany}=x$
$sec^2y\frac{d}{dx}=1$
$\frac{d}{dx}=cos^2y$

Don't see how $cos^2y = \frac{1}{1+x^2}$

4. ## Re: product of inverse trig function

Originally Posted by Barthayn
I understand I have to use the product rule, but I don't see how $\frac{d[\arctan(x)]}{dx}=\frac{1}{1+x^2}$.
I get ${tany}=x$
$sec^2y\frac{d}{dx}=1$
$\frac{d}{dx}=cos^2y$
Don't see how $cos^2y = \frac{1}{1+x^2}$
$\sec^2(y)=1+\tan^2(y)=1+x^2$

5. ## Re: product of inverse trig function

Originally Posted by Plato
$\sec^2(y)=1+\tan^2(y)=1+x^2$
Thank you so very much! I was on this problem for a long time yesterday and today and I never thought to use secy where it was I always used cosy and got stuck so many times! Thank you!