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About Scotto
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Any kind of mathematics (calculus, analysis, game theory, linear approximation, finite differences, linear regression, linear programming, numerical analysis, probability, statistics, etc.). I also have answered some questions in Physics (mass, momentum, falling bodies), Chemistry (charge, reactions, symbols, molecules), and Biology.

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Experience in the area: I have tutored students in all areas of mathematics for over 20 years. Education/Credentials: BSand MS in Mathematics from Oregon State University, where I completed sophomore course in Physics and Chemistry. I received both degrees with high honors. Awards and Honors: I have passed Actuarial tests 100, 110, and 135.

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Maybe not a publication, but I have respond to well oveer 3000 questions on the PC. That's around 2,000 in basic math and 1,000 in advanced math.

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I aquired well over 40 hours of upper division courses. This was well over the number that were required. I graduated with honors in both my BS and MS degree from Oregon State University. I was allowed to jump into a few junior level courses my sophomore year.

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I have been nominated as the expert of the month several times. All of my scores right now are at least a 9.8 average (out of 10).

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You are here: Experts > Teens > Homework/Study Tips > Calculus > zeroes

Calculus - zeroes

Expert: Scotto - 11/6/2009

Question
I have found the derivative of this function i asked about earlier but now i can't seem to find a way to solve for the zeroes because i cant isolate x after simplifying the derivative.

f(x) = (2009^x)/(x^2009) where x > 0

The simplified derivative that i got was:

f'(x) = 2009^x[ln2009(x^2009) + 2009x^2008]

Answer
The derivative is the quotient rule where f(x) = g(x)/h(x) where
g(x) = 2009^x and h(x) = x^2009.

g'(x) = ln(2009)2009^x and h'(x) = 2009x^2008.

The derivative is (lo d hi - hi d lo)/lo². This gives us
f'(x) = (x^2009*ln(2009)2009^x - 2009^x*2009x^2008)/x^4018.

Note that 2008^x can be taken out of each term in the numerator and denominator, giving
f'(x) = (x*ln(2009)2009^x - 2009^x*2009)/x^2010.

Factoring out 2009^x gives f'(x) = 2009^x(x*ln(2009) - 2009)/x^2010.

That's what I get, and you're close.

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