Posted by Soroban on August 26, 2002 at 13:17:50:
In Reply to: Re: second derivative posted by bana on August 25, 2002 at 19:15:20:
: Thank you for your help.
: I tried it out using the product rule and quotient rule, but I'm stuck again because what I got at the end seems a bit strange.
: For this question y=4x^2Ln(x+5)
: First I got
: y'=8x.Ln(x+5)+ 4x^2.1/(x+5)
This is correct. So, the first derivative is:
y' = 8x*ln(x+5) + 4x^2/(x+5)
: y'=8.1/(x+5) + 8x.-1(x+5)^-2
: y'=8/(x+5) - 8x/(x+5)^2
But what did you do here, Bana?
It looks like you were taking a derivative
(incorrectly, by the way), in which case, it
would be the second derivative, y", at this
point. (You didn't use the Product Rule on the
first term, nor the Quotient Rule on the second
term - the fraction.)
So the rest is wrong, too. You didn't know it,
but you were working on the THIRD derivative.
: But then in the second derivative,I got
: y"=[(0(x+5)-8(1))/(x+5)^2] - [(8(x+5)^2 - 2(x+5).1)}/(x+5)^4]
Let's start at the first derivative:
y' = 8x*ln(x+5) = 4x^2/(x+5)
The first term is a Product: its derivative is
8*ln(x+5) + 8x/(x+5)
The second is a Quotient: its derivative is
[(x+5)*8x - 4x^2*1]/(x+5)^2 = (4x^2+40x)/(x+5)^2
So, y", the second derivative, is ALL of that:
y" = 8*ln(x+5) + 8x/(x+5) + (4x^2+40x)/(x+5)^2
Basically, we're done - unless they want us to
get a common denominator and make one fraction
out of it.
y" = [8(x+5)^2*ln(x+5) + 12x^2 + 80x]/(x+5)^2
*whew* I hope one of those two answers shows up
in the Answers.