Question

Find the indicated derivative of the following

Answer 1

\[\frac{ d}{ dx} ((\sqrt[5]{x})^{\ln(x))}\]

Answer 2

PLease tell me it just looks more complicated than it really is

Answer 3

chain rule?

Answer 4

\[y=(\sqrt[5]{x})^{\ln(x)} \\ \ln(y)=\ln(x) \ln(\sqrt[5]{x})\]
Now try to find the derivative of both sides

Answer 5

its called log differentiation
since I took log of both sides
we will find y' implicitly now

Answer 6

\[\frac{ 1 }{ y }*y' = \frac{ 1 }{ x } * 5 \ln \frac{ 1 }{ \sqrt{x} }\]

Answer 7

you need product rule on the right hand side

Answer 8

right right, okay.

Answer 9

\[\frac{1}{y} y'=(\ln(x))'\ln(x^\frac{1}{5})+\ln(x)(\frac{1}{5} \ln(x))' \]

Answer 10

by the way
\[\ln(\sqrt[5]{x})=\ln(x^\frac{1}{5})=\frac{1}{5}\ln(x)\]

Answer 11

thanks for the tip! so now i think I can factor the "ln(x)" correct?

Answer 12

well have you found (ln(x))' and (ln(x)/5)' yet?

Answer 13

I think you know (ln(x))'=1/x from above

Answer 14

so (c ln(x))'=c(ln(x))'=c*1/x=c/x
where c is a constant so you should know how to different the (1/5*ln(x)) part

Answer 15

differentiate*

Answer 16

okay so (1/5*ln(x))' = 1/5* 1/x

Answer 17

right so you have this
\[\frac{1}{y} y'=(\ln(x))'\ln(x^\frac{1}{5})+\ln(x)(\frac{1}{5} \ln(x))' \\ \frac{y'}{y}=\frac{1}{x}\ln(x^\frac{1}{5})+\ln(x) \frac{1}{5} \frac{1}{x}\]

Answer 18

we can make this a little prettier

Answer 19

and then multiply both sides by y

Answer 20

y' = \[y' = (\frac{ \ln x ^{\frac{ 1 }{ 5 }} }{ x } + \frac{ \ln(x) }{ 5x}) * y\]

Answer 21

\[\frac{y'}{y}=\frac{1}{5} \frac{1}{x}\ln(x)+\frac{1}{5}\frac{1}{x}\ln(x)\]
well and you can bring that 1/5 down

Answer 22

so you know a+a=2a

Answer 23

combine the like terms?

Answer 24

yeah

Answer 25

y' = 2/5+ 2/x ln(x)^2

Answer 26

\[y'=\frac{2 \ln(x)}{5x} y\]

Answer 27

oh, I get what your are saying! Thanks a done, sorry for being terrible at calc

Answer 28

ton*

Answer 29

you really weren't that bad at the calculus part
once I said you needed to use the product rule