Question

Hi, how would you solve...

Answer 1

first step: take ln for both sides and apply
\[\ln\frac{ a }{ b }=lna-lnb\]

Answer 2

@esamalaa okay, what would be the next step?

Answer 3

there's another ln rule says
\[\ln(a*b)=lna+lnb\]
look to the first term in the right hand side and apply this rule
maybe you need to apply this rule too
\[\ln(a)^{b}=b*lna\]

Answer 4

all these steps are only for simplify the problem to be ready fir derivative

Answer 5

okay I understand that, but what do we do with the ln(y) after applying the above steps? Do I just differentiate the RHS and then put it to the power of e to cancel the ln(y)?

Answer 6

no you have to differentiate the LHS too
\[lny \rightarrow \frac{ 1 }{ y }*\frac{ dy }{ dx }\]

Answer 7

Is it okay if you can send me like a partial solution or something? I'm still quite confused... and thanks so much for the help :)

Answer 8

that is a very disgusting function haha

Answer 9

hahaha tell me about it

Answer 10

\[y=\frac{ 3^xx ^{\sqrt{x}} }{\sqrt[3]{x^4+2x} }\]
\[=\frac{ 3^xx \sqrt{x} }{ \left( x^4+2x \right)^{\frac{ 1 }{ 3 }} }\]
\[\ln y=\ln \frac{ 3^xx ^{\sqrt{x }} }{ \left( x^4+2x \right)^{\frac{ 1 }{ 3 }} }\]
\[=\ln \left( 3^xx ^{\sqrt{x}} \right)-\ln \left( x^4+2x \right)^{\frac{ 1 }{ 3 }}\]
\[=\ln 3^x+\ln x ^{\sqrt{x}}-\frac{ 1 }{ 3 }\ln \left( x^4+2x \right)\]
\[=x \ln 3+\sqrt{x}\ln x-\frac{ 1 }{ 3 }\ln \left( x^4+2x \right)\]
now you can differentiate.

Answer 11

Ohhh thanks so much @esamalla @surjithayer I get it now!!

Answer 12

yw