Help Please! Attachments to come...

Question

poofypenguin · Answer

Here is the question:

poofypenguin · Answer

Here is the work i've done so far:

poofypenguin · Answer

The answer is incorrect, and I'm not sure what i did wrong!

anonymous · Answer

Okay, So you made a little mistake in your derivative.

anonymous · Answer

Instead of a 5, It should be 10

tkhunny · Answer

Right off the top.  $$ln\left(5\cdot x^{ln(x)}\right) = ln(5) + ln\left(x^{ln(x)}\right)$$

poofypenguin · Answer

@zordoloom which line is this mistake on...I'm completely blind! lol

anonymous · Answer

attachment

tkhunny · Answer

As soon as you introduced the logarithm.  That "ln(x)" is NOT an exponent on the '5' - only on the 'x'.

poofypenguin · Answer

@tkhunny Ooooooh! I get it now! Let me try it!

poofypenguin · Answer

So far i've gotten $$\ln 5x ^{\ln x} = \ln (5) + [\ln (x)]^{2}$$

if this is correct, is there an easier way to differentiate the ln(x) ^2 without using the product rule?

poofypenguin · Answer

@tkhunny are you still there?

poofypenguin · Answer

anyone?

anonymous · Answer

yes

anonymous · Answer

don't forget 
ln (y)  = ln(5) +(ln(x))^2

poofypenguin · Answer

alright, but is there a trick to doing the ln(x) ^2 without using the product rule?

anonymous · Answer

mm  no trick needed :)

anonymous · Answer

2*ln(x) * 1/x

just the chain rule

anonymous · Answer

got it now?

anonymous · Answer

A mathematica presentation is attached.  Log in Mathematica is Ln

poofypenguin · Answer

@Algebraic! Yep I got it now! Thanks so much! :D

anonymous · Answer

sure:)