If y=(x-5) / (lnx), find y'(e^2).

Question

The answer is ((e^2)+5) / (4e^2) but I am not sure how to get to it.

Answer 1

Use quotient rule to differentiate as follows:

y = (x - 5) / lnx

y^' = [1·lnx - 1/x·(x - 5)] / ln²x

y^'(e²) = [1·lne² - 1/e²·(e² - 5)] / ln²e²

= [2 - 1 + 5/e²)] / 4

= [e² + 5] / 4e²

Answer 2

Using the quotient rule, y' = [(lnx)(1) - (1/x)(x-5)] / [lnx]²

Replacing x by e², we have y' = [(2)(1) - (1/e²)(e²-5)] / 4 (Recall that ln(e^x) = x)

So, y' = (1 + 5/e²) / 4 = (e² + 5) / (4e²)