Jt061952 SG I am having difficulty differentiating this problem. Can someone kindly help?Differentiate: ln ( x + sqrt (x^2 - 1))The answer given in the book is given as: 1 / sqrt (x^2 - 1). Thank you.
wee_ws SG Hi Jt061952, Let y = ln [x + sqrt(x^2 - 1)]. Then dy/dx = [ 1 + (2x)/sqrt(x^2 - 1) ] / [ x + sqrt(x^2 - 1) ], which simplifies to your said answer. In general, dy/dx = f'(x) / f(x) if y = ln f(x). Thanks!Cheers, Wen Shih
Forbiddensinner SG Edited : Drats. I am 100% sure now I made a mistake somewhere.My apologies to Lecturor Wee.On the otherhand, do note that it is[ 1 + x/sqrt(x^2 - 1) ] / [ x + sqrt(x^2 - 1) ]rather than[ 1 + (2x)/sqrt(x^2 - 1) ] / [ x + sqrt(x^2 - 1) ]
wee_ws SG Hi, You may wish to check with Wolfram Integrator at: http://integrals.wolfram.com/index.jsp Wolfram indeed returns the answer ln [x + sqrt(x^2 - 1)]. Thanks!Cheers, Wen Shih
Jt061952 SG WS, Initially I tried using the chain rule as in Wolfram and it got horrendous and the answer did not come out in the simplified form. But will check again. Thanks again.
wee_ws SG Hi, Just type in 1/(Sqrt[x^2 - 1]) and the answer comes out in an instant. Anyhow, my working above is correct. Thanks!Cheers, Wen Shih