Question

Differentiation & Integration help. *question attached below*

Answer 1

No Q

Answer 2

Can I have some help on this question, please? I've been stuck on it for the past 4 hours with no luck

Answer 3

What the derivative of inverse tan

Answer 4

The derivative of inverse tan is
\[\frac{ 1 }{ 1+x ^{2} }\]

Answer 5

Have you learned `Integration by Parts` yet? :)

Answer 6

Yes, but not in depth

Answer 7

So you deterimined:\[\Large\rm \frac{d}{dx}\tan^{-1}(\color{royalblue}{x})=\frac{1}{1+(\color{royalblue}{x})^2}\]

Our function will follow the same pattern, but we'll also need to apply the chain rule.

Answer 8

So here is how we would set it up,
\[\large\rm \frac{d}{dx}\tan^{-1}(\color{royalblue}{\sqrt{x^2-1}})=\frac{1}{1+\left(\color{royalblue}{\sqrt{x^2-1}}\right)^2}\color{royalblue}{\left(\sqrt{x^2-1}\right)'}\]

Answer 9

Same idea, argument goes into that blue spot.
But we chain rule, yes?

Understand how to take the derivative of that square root? :o

Answer 10

We raise it to the 1/2 power?

Answer 11

Ummm yah you can do that if you prefer to use power rule.
This is a good derivative to memorize though, it shows up a lot.\[\Large\rm \frac{d}{dx}\sqrt x=\frac{1}{2\sqrt x}\]

Answer 12

Hmm, okay

Answer 13

\[\rm \frac{d}{dx}\tan^{-1}(\color{royalblue}{\sqrt{x^2-1}})=\frac{1}{1+\left(\color{royalblue}{\sqrt{x^2-1}}\right)^2}\left(\frac{1}{2\sqrt{x^2-1}}\right)\color{royalblue}{(x^2-1)'}\]So we get our 1 over 2 square roots when we differentiate.
Then you have to chain rule again.

Answer 14

Why do we have to chain rule again?

Answer 15

Because...? :o

Answer 16

If chaining ever gives you something more than just 1, then you have to apply it.

Example:\[\Large\rm \frac{d}{dx}(x)^2=2(x)\frac{d}{dx}(x)\quad=\quad 2(x)(1)\]Applying chain rule produced a 1, which was insignificant.
So you don't need to apply the chain rule in that example.

But when we have something like:\[\Large\rm \frac{d}{dx}(x^2-1)^2=2(x^2-1)\frac{d}{dx}(x^2-1)\]\[\Large\rm =2(x^2-1)(2x)\]See how our chain produced more than just a 1?
So it was needed.

Answer 17

Oh, yes