Question

derivative of this

Answer 1

\[\ln (x{+}\sqrt{x^2-1})\]

Answer 2

@zepp

Answer 3

You'll need to apply the chain rule to this, can you identify the outter function?

Answer 4

ln

Answer 5

Good, what is the derivative of ln(x)?

Answer 6

\[\frac{ 1 }{ x }\]

Answer 7

Excellent, now, let's identify this outter function ln(x) as f(x) and this inner function as g(x)
Applying the chain rule, we get \[\frac{d}{dx}f(g(x))=f'(g(x))g'(x)\]

Answer 8

We got the first part: \(f'(g(x))\), which is the derivative of \(\ln(g(x))\), \(\large \frac{1}{g(x))}\)

Answer 9

Still following? :P

Answer 10

I am following. but, couldnt I just use the chain rule variant of \[\frac{ d }{ dx }(\ln[f(x)])=\frac{ f'(x) }{ f(x) }\]?

Answer 11

Absolutely! And that'll be our second step, since there's another function inside the natural logarithm function.

Answer 12

ok. so then here is where i am, \[-f'(x) = \frac{1 + \frac{ 1 }{ 2 }(x-2)^-(\frac{ 1 }{ 2 } }{x+\sqrt(x^2-1) }\]

Answer 13

*** ignore that - in front of f(x)

Answer 14

Yes, looks like, but the problem is that there's another function inside the square root function, so you'll need to differentiate it as well :P

Answer 15

the one in the denominator?

Answer 16

\[\large \frac{d}{dx}\ln (x{+}\sqrt{x^2-1})=\frac{\frac{d}{dx}(x+\sqrt{(x^2-1)})}{x+\sqrt{x^2-1}}\]

Answer 17

Uhh, the x^2-1, you'll need to differentiate this. What you did was great, but there was only this one that was missing :D

Answer 18

\[\large f'(x) = \frac{(1 + \frac{ 1 }{ 2 }(x^2-2)^{-\frac{ 1 }{ 2 }} )(\frac{d}{dx}(x^2-2))}{x+\sqrt(x^2-1) }\]

Answer 19

According to the chain rule, leave the inner function intact and perform the derivative operation on the outter function, then multiply what you got by the derivative of the function that you left intact.

Answer 20

so this would be it \[\large f'(x) = \frac{(1 + \frac{ 1 }{ 2 }(x^2-2)^{-\frac{ 1 }{ 2 }} )(2x)}{x+\sqrt(x^2-1) }\]

Answer 21

Yup! :D