Question

Find the second derivative of the function.
g(x) = 5sqrt[x] + e^(2x) ln(x)

y"(x)=

Answer 1

thanks for ur previous help:)i appreciate it

Answer 2

\[g=5x^\frac{1}{2}+e^{2x} \ln(x)\]
\[g'=5 \cdot \frac{1}{2} x^{\frac{1}{2}-1}+2e^{2x} \ln(x)+e^{2x} \cdot \frac{1}{x}\]

Answer 3

\[g'=\frac{5}{2}x^\frac{-1}{2}+e^{2x}(2\ln(x)+\frac{1}{x})\]
now we to do the derivative again
\[g''=\frac{5}{2} \cdot \frac{-1}{2} x^{\frac{-1}{2}-1}+2e^{2x}(2 \ln(x)+\frac{1}{x})+e^{2x}(2 \frac{1}{x}-\frac{1}{x^2})\]

Answer 4

np obsolete

Answer 5

myininaya for step five with the second derivative shouldn't it be -5/4

Answer 6

because 2 times 2 is 4 lol

Answer 7

yes

Answer 8

\[g''=\frac{-5}{4\sqrt{x^3}}+e^{2x}(4 \ln(x)+2\frac{1}{x}+2\frac{1}{x}-\frac{1}{x^2})\]
\[g''=\frac{-5}{4\sqrt{x^3}}+e^{2x}(4 \ln(x)+\frac{4}{x}-\frac{1}{x^2}) \]

Answer 9

myininaya your hands are going to fall off ;)

Answer 10

lol

Answer 11

yes sorry just pointing that out and also \[\sqrt{x ^{3}}= x \sqrt{x}\]

haha @ Zarkon

Answer 12

But that's just simplification

Answer 13

i think i won't simplify anymore just because i forget to do stuff like multiply 2 times 2

Answer 14

Yeah...just glad we caught that...algebra really sucks when you taking the derivative one small thing can mess u up lol

Answer 15

hmm im really glad u guys r helping me out so what woud the the final answer be:S

Answer 16

like if you ould just point out where u put it

Answer 17

IT should be myininaya's fourth post after mine..In that post the second answer is the simplified one.

Answer 18

I mean 3rd post

Answer 19

hmm thanks so much, thats awfully long.r u sure its right?maybe this is one of the long oness

Answer 20

?

Answer 21

Believe me I've down worse questions ones than this problem...it's right. What I saw was just small error. Of course when you are typing sometimes you make mistakes too.

Answer 22

derivatives
i just posted my question. im stuck in the middle of it .

Answer 23

chaise can u please read my question i posted it already