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What Is The Nth Derivative Of Ln(Y+C)?

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StarDiwakar | 06:58 Wed 06th Aug 2014 | Science
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If c is a constant and y is a function of x, then what is the nth derivative of ln(y+c)?
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For this you'll need to use the Chain Rule once and then Leibniz's Rule on the resulting first derivative. The first step then gives that: (d/dx)(ln(y+c)) = (dy/dx) / (y+c) which is a result that you should be able to produce without difficulty. There is no easy way to write out the general form of the nth derivative in this Answerbank text box but, using Leibniz...
07:33 Wed 06th Aug 2014
For this you'll need to use the Chain Rule once and then Leibniz's Rule on the resulting first derivative. The first step then gives that:

(d/dx)(ln(y+c)) = (dy/dx) / (y+c)

which is a result that you should be able to produce without difficulty. There is no easy way to write out the general form of the nth derivative in this Answerbank text box but, using Leibniz rule that applies to the nth derivative of the product u(x)*v(x) for two functions u, v, you will be able to get the answer by letting u(x) = dy/dx and v(x) = 1/(y+c), both of which are easily differentiable.
great JIm nice to seeing you contributing on [6y below ] your specialist subject !

I hope inquiry this finds you well
42
venator if you cant make a sensible contribution, to differential calculus - A level or a bit above,

then stick to chadderbank !
You could write ln(y+c) as ln (c+y) =ln c + ln(1 + y/c) and then expand the latter using the Taylor series, then find the nth derivative
For any large value of y, though, that will make no sense -- and since we aren't told what the value of c is, or what values y can take, I'd stray away from doing it that way. The approach I outlined above is completely general and valid for all y and c as long as their sum is positive.
Okay, thanks jim. I was just thinking aloud.
Could we use a simple substitution u= y+c and then the chain rule to find dx/dy:
dx/dy= dx/du*du/dy which, without pen and paper, I think is simply 1/u which is 1/(y+c).
I think it's possible to differentiate that repeatedly and see the pattern to emerge, although I think there will be factorials. When I get time I'll have a look, although I don't have the patience I used to have for these things 35-40 years ago. Or maybe I'll just wait and see the answer you or Star come up with!

I'd considered completing the calculation but even supposing I could be bothered to get the general result I'm not sure it would be that enlightening, and rather difficult to present here anyway. I've presented the method and the answer for the first derivative, so it's up to Star to use that to get to the full final answer. Whether or not it will have a simple form, I don't know. On the face of it I suspect that the answer won't be too exciting, since after all the question posed is for a very general function of x.

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