Question

help on derivatives pls

Answer 1

looks like you did the product rule but didn't apply properly.

if you use the product rule, \(v=\left(\ln x \right)^{-1}\)

Answer 2

\[G \left( x \right)=\frac{ x }{ \ln x }\]yes?

Answer 3

yep

Answer 4

and you want to find\[G\,'\left( e \right)\]

Answer 5

indeed

Answer 6

use the quotient rule... you have u' and v' correct, just need to put things together in the right way

Answer 7

i don't really see where i wen't wrong cause i thought lnx = 1/x ??

Answer 8

\[\frac{ d }{ dx }\left( \ln x \right)=\frac{ 1 }{ x}\]
\[\text{If }y=\frac{u \left( x \right)}{ v \left( x \right)} \text{, then }y'=\frac{ u' \cdot v-u \cdot v' }{ v^2 }\]

Answer 9

oh sugar i used the wrong rule! so does it become lnx-1/lnx^2 ??

Answer 10

\[G\,'\left( x \right)=\frac{ \left[\frac{ d }{ dx }\left( x \right) \right]\ln x-x \left[ \frac{ d }{ dx }\left( \ln x \right) \right]}{ \left( \ln x \right)^2 }\]

Answer 11

yes, if in parentheses (ln x - 1)/(ln x)^2

Answer 12

when i plugged in e i got 0, is that right?

Answer 13

and so G'(e) = o, correct!

Answer 14

0, not o.

Answer 15

good to go?

Answer 16

yep! just one last question lol, is this right?

Answer 17

is it\[k \left( x \right)=\sin^{-1} \left( \frac{ 1+x }{1+2x } \right)\]

Answer 18

yep!:)

Answer 19

well then you totally forgot about the inverse sin!

Answer 20

\[\frac{ d }{ dx }\left[ \sin^{-1} \left(f \left( x \right) \right) \right]=\frac{ f ' \left( x \right) }{ \sqrt{1-f \left( x \right)^2} }\]you found the numerator only...

Answer 21

lol shoot... okay so i redid it and subbed 2 in and got suqare root 3 over 6?

Answer 22

\[k\,'\left( 2 \right)=\frac{ -\frac{ 1 }{25 } }{ \sqrt{1-\frac{ 9 }{ 25 }} }=\frac{ -\frac{ 1 }{ 25 } }{ \sqrt{\frac{ 16 }{ 25 } }}=\frac{ -\frac{ 1 }{ 25 } }{ \frac{ 4 }{ 5 }} =-\frac{ 1 }{ 25 }\cdot \frac{ 5 }{ 4 }\]

Answer 23

\[k\,'\left( x \right)=\frac{ -\frac{ 1 }{\left( 1+2x \right)^2} }{ \sqrt{1-\left( \frac{ 1+x }{ 1+2x } \right)^2} }\]

Answer 24

huh @pgpilot326 i don't get that at all?