Question

Find f'(x) for f(x)=ln(x+rad(x^(2)-1)). Can somebody please help me with this?

Answer 1

\(\dagger\)..

Answer 2

Almost, but what you've written is kind of ambiguous in the first step of writing out the derivative. \[f'(x) \ne \frac{d}{dx}(\ln(x+\sqrt{1+x^2}))*\frac{d}{dx}(x+\sqrt{1+x^2})\] \[f'(x) = \frac{d}{dx}(\ln(x+\sqrt{1+x^2}))\]

you've sort of combined the chain rule there where it shouldn't be.

Answer 3

Oh, I don't think I was een done yet. If you combine some things in my last step it could simplify even further.

Answer 4

So to put it explicitly, \[f(x)= h(k(x))\] \[h(x)=\ln(x) \\ k(x) = 1+\sqrt{1+x^2}\] \[f'(x)=h'(k(x))*k'(x)\]
\[h'(x) = \frac{1}{x} \\ h'(k(x))=\frac{1}{1+\sqrt{1+x^2}} \\ k'(x) = \frac{x}{\sqrt{1+x^2}}\]

Answer 5

Omg, i also realized I mixed up a lot of my signs :|

Answer 6

\[\color{red}{\sf f(x) = \ln(x+\sqrt{x^2-1})}\]Just pretend the derivative i am taking to the right is actually a fraction ontop of the derivative of the log function.\[\sf f'(x) = \frac{d}{dx}(\ln(x+\sqrt{x^2-1}) \cdot \frac{d}{dx} (x+\sqrt{x^2-1})\]\[\sf f'(x) = \frac{1}{x+\sqrt{x^2-1}}\cdot \left(1+\frac{1}{2}(x^2-1)^{-1/2}\cdot 2x\right)\]\[\sf f'(x)= \left[\frac{1}{x+\sqrt{x^2-1}}\right]\cdot \left(1+\frac{x}{(x^2-1)^{1/2}}\right)\]\[\sf f'(x)=\left[\frac{\left(1+\frac{x}{\sqrt{x^2-1}}\right)}{\sqrt{x^2-1} \left(\frac{x}{\sqrt{x^2-1}}+1\right)}\right]\]Realize, once again, that i just took the numerator and separated it from the denominator at first to simplify it for myself, but I just put it back ontop.If we follow the format \(\frac{a}{b(a)} =\frac{1}{b}\) we will end up with :\[\sf \boxed{\large f'(x) = \frac{1}{\sqrt{x^2-1}}}\] :)

Answer 7

\[\sf f'(x)= \left[\frac{1}{x+\sqrt{x^2-1}}\right]\cdot \left(1+\frac{x}{\sqrt{x^2-1}}\right)\\
=\left[\frac{1}{x+\sqrt{x^2-1}}\right]\cdot \left(\frac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}}\right)\\
=\frac{1}{\sqrt{x^2-1}}\]

Another sorter method by @UnkleRhaukus :)

Answer 8

\(\color{blue}{\text{Originally Posted by}}\) @Jhannybean
\[\color{red}{\sf f(x) = \ln(x+\sqrt{x^2-1})}\]Just pretend the derivative i am taking to the right is actually a fraction ontop of the derivative of the log function.\[\sf f'(x) = \frac{d}{dx}(\ln(x+\sqrt{x^2-1}) \cdot \frac{d}{dx} (x+\sqrt{x^2-1})\]
\(\color{blue}{\text{End of Quote}}\)

Don't write it if it's wrong. That's wrong. Calculus is confusing enough.

Answer 9

Asked a number of my math professors about the format on this problem, and was told there was nothing wrong with writing it that way.

Answer 10

Some messages look like they are deleted/missing for me. Kainui, was your reply to a post that the OP deleted?

Answer 11

PS is this mathjax or what even is this? This doesn't look like LaTeX in its typical format. Is this just upright math LaTeX 24/7 or a different font or something?

Answer 12

It's just a different font. `/sf`

Answer 13

Hmm.. maybe it would have been better if I let \(u=x+\sqrt{x^2-1}\) and \(du=1+\frac{x}{\sqrt{x^2-1}}\) Or something. Oh freaking well.

Answer 14

It's really bothering me now.

Answer 15

I know it's somewhere in my Calc book, there's a better way to do what you're trying to illustrate using composite functions, but I haven't looked it up yet; might eventually.

Answer 16

You mean conjugates?

Answer 17

Hmm...

Answer 18

I think what you SHOULD have written was: \[
f'(x)=\frac{d}{d\left(x+\sqrt{1+x^2}\right)}\ln\left(x+\sqrt{1+x^2}\right)\cdot \frac{d}{dx}\left(x+\sqrt{1+x^2}\right)

\]Or if you had said \(u = \left(x+\sqrt{1+x^2}\right)\) then: \[
f'(x)=\frac{d}{du}\ln\left(u\right)\cdot \frac{d}{dx}\left(x+\sqrt{1+x^2}\right)
\]

Answer 19

This ultimately would mean: \[
\frac{df}{dx}=\frac{df}{du}\frac{du}{dx}
\]However, it's a minor notation error.
What you were trying to say was correct, but what you actually wrote was a bit incorrect.

Answer 20

Technically: \[\begin{split}
f'(x)&=\frac{d}{dx}\ln\left(x+\sqrt{1+x^2}\right)\cdot \frac{d}{dx}\left(x+\sqrt{1+x^2}\right) \\
&= f'(x)\frac{d}{dx}\left(x+\sqrt{1+x^2}\right)
\end{split}
\]Which doesn't make sense.

Answer 21

This makes more sense. Thank you.