1. ## differentiating arctan

Hi there,

I came across a rule that says the derivative of arctanx is $1/(1 + x^2)$. How would you differentiatie arctanx to get this?

Also, is there a way to show fractions with one factor on top of another, rather than side by side as in my example?

Thanks,

jjmclell

2. $y = \arctan x \implies \tan y = x \implies y'\sec ^2 y = 1.$

Can you take it from there?

3. Let $f(x) = \tan^{-1} x$ this function is differenciable on $\mathbb{R}$. Let $g(x) = \tan x \mbox{ on }(-\pi/2,\pi/2)$ which is also differenciable. Then, $g(f(x))=x$. Using the chain rule $f'(x)g'(f(x))=1$ thus $f'(x)\sec^2 (\tan^{-1} x)=1$ thus $f'(x)(1+x^2) = 1\implies f'(x) = \frac{1}{1+x^2}$.

4. Originally Posted by jjmclell
Hi there,

I came across a rule that says the derivative of arctanx is $1/(1 + x^2)$. How would you differentiatie arctanx to get this?

Also, is there a way to show fractions with one factor on top of another, rather than side by side as in my example?

Thanks,

jjmclell
Put $u=\arctan(x)$ , then $\tan(u)=x$, so:

$\frac{d}{dx}\tan(u)=\sec^2(u) \frac{du}{dx}=1$

and from this point its just algebra and trig identities.

RonL

5. Originally Posted by CaptainBlack
Put $u=\arctan(x)$ , then $\tan(u)=x$, so:

$\frac{d}{dx}\tan(u)=\sec^2(u) \frac{du}{dx}=1$

and from this point its just algebra and trig identities.

RonL

Ok, I clearly see that if $u=arctan(x)$ then $tan(u)=x$. However, why are we suddenly doing $\frac{d}{dx}\tan(u)$? Aren't we interested in $\frac{du}{dx}$?

6. Originally Posted by jjmclell
Ok, I clearly see that if $u=arctan(x)$ then $tan(u)=x$. However, why are we suddenly doing $\frac{d}{dx}\tan(u)$? Aren't we interested in $\frac{du}{dx}$?
Look at what you get when you do that differentiation (not difficult I have done it for you)

RonL

7. Originally Posted by CaptainBlack
Put $u=\arctan(x)$ , then $\tan(u)=x$, so:

$\frac{d}{dx}\tan(u)=\sec^2(u) \frac{du}{dx}=1$

and from this point its just algebra and trig identities.

RonL

Ok, I see that we wind up with $\sec^2(u)\frac{du}{dx}$ which contains $\frac{du}{dx}$ which we can isolate if we set it equal to 1. I even see how we get the 1 since $\tan(u) = x$ and $\frac{dx}{dx} = 1$, but how does $\frac{d}{dx}\tan(u) = sec^2\frac{du}{dx}$? I know the derivitave of $\tan(u) = sec^2(u)$ but that doesn't explain why it's multiplied by $\frac{du}{dx}$ in this example. Sorry, I haven't taken calculus since highschool and now I'm trying to relearn it.

8. Originally Posted by jjmclell
Ok, I see that we wind up with $\sec^2(u)\frac{du}{dx}$ which contains $\frac{du}{dx}$ which we can isolate if we set it equal to 1. I even see how we get the 1 since $\tan(u) = x$ and $\frac{dx}{dx} = 1$, but how does $\frac{d}{dx}\tan(u) = sec^2\frac{du}{dx}$? I know the derivitave of $\tan(u) = sec^2(u)$ but that doesn't explain why it's multiplied by $\frac{du}{dx}$ in this example. Sorry, I haven't taken calculus since highschool and now I'm trying to relearn it.
Its the chain rule:

$\frac{d}{dx}f(u(x))=u'(x) ~f'(u(x))$

Now in your case $f=\tan$, so you
have:

$
\frac{d}{dx}\tan(u)=\sec^2(u) \frac{du}{dx}
$

RonL

9. Ahh, I see. So when you're taking a trig function of another trig function, say as in this case where we have $\tan(\arctan(x))$, is this always considered a composite function?

10. Originally Posted by jjmclell
Ahh, I see. So when you're taking a trig function of another trig function, say as in this case where we have $\tan(\arctan(x))$, is this always considered a composite function?
Yes, if you looked at post #3 that was exactly what I was doing.

11. Wow, it took me a while to figure out how $\sec^2(u) = 1+x^2$ but all is now clear. Thanks for everyone's help.