Few Problems (Derivatives)

1. Nov 13, 2006

TommyLF

I need a little help on the following problems:

Find the 2nd derivative at the indicated point:
y= x(ln x) At the point (1,0)

I got to:

y''= $$\frac{1}{x}$$

So do I just sub in 1 for x? And then the answer would be 1?

Also, I need help on the following problem:

Evaluate the first derivative at the given value X: x=0

[4^(3x)(x^3 - x + 1)^(1/5)] (And then all of that raised to the -2)
-------------------------
(x^2 + x + 1)^(4)

2. Nov 13, 2006

JasonRox

Yeah, basically put 1 in the equation for the first question.

For the second one, I see no trick. I guess you have to painfully use the Quotient Rule, Chain Rule and Product in the right order. Have fun doing that. :S

Actually, I'd probably do it using first principles. It seems like the easier way out. (First principles is finding the derivative using the limit definition.)

3. Nov 14, 2006

HallsofIvy

Staff Emeritus
JasonRox, you have a wicked sense of humor! Anyone trying to find the derivative of that second function using the limit of the difference quotient is guarenteed to go insane!

TommyLF, I would recommend going ahead and incorporating that -2 exponent into the numerator and, because that is a negative exponent, writing the whole thing as a product of terms with negative exponents:
$$\frac{[4^{3x}(x^3- x+ 1)^{1/5}]^{-2}}{(x^2+ x+ 1)^4}= 4^{-6x}(x^2-x+1)^{-2/5}(x^2+ x+ 1)^{-4}$$
and use the product rule and chain rule.