ln differentiation

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kendrick - 22 Jul 2006 15:37 GMT

pls, can somebody help me solve this problem

differentiate the following function

y=ln 5/x

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Frederick Williams - 22 Jul 2006 15:49 GMT

> pls, can somebody help me solve this problem
>
> differentiate the following function
>
> y=ln 5/x

y = log(5/x) = log 5 - log x

dy/dx = -1/x.

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G.E. Ivey - 22 Jul 2006 16:57 GMT

Fredrick Williams did it the easy way. Here's the hard way just to show it gives the same answer:

y= ln(5/x)= ln(u) where u= 5/x.

dy/du= 1/u and du/dx= (0*x- 5*1)/x^2= -5/x^2 (by the quotient rule) or u= 5x^(-1) so du/dx= -5/x(-1-1)= -5x^(-2)= -5/x^2.

By the chain rule, then dy/dx= (dy/du)(du/dx)= (1/u)(-5/x^2)= (1/(5/x))(-5/x^2)= (x/5)(-5/x^2)= -1/x.

Yes, y= ln(5/x)= ln(5)- ln(x) so dy/dx= -1/x is much easier!

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Math Forum / Mathematics / Undergraduate Math / July 2006

ln differentiation