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FireKat97
 one year ago
Hi, how would you solve...
FireKat97
 one year ago
Hi, how would you solve...

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0first step: take ln for both sides and apply \[\ln\frac{ a }{ b }=lnalnb\]

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0@esamalaa okay, what would be the next step?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there'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 \[\ln(a)^{b}=b*lna\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0all these steps are only for simplify the problem to be ready fir derivative

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0okay 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)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no you have to differentiate the LHS too \[lny \rightarrow \frac{ 1 }{ y }*\frac{ dy }{ dx }\]

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0Is 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 :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is a very disgusting function haha

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0hahaha tell me about it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[y=\frac{ 3^xx ^{\sqrt{x}} }{\sqrt[3]{x^4+2x} }\] \[=\frac{ 3^xx \sqrt{x} }{ \left( x^4+2x \right)^{\frac{ 1 }{ 3 }} }\] \[\ln y=\ln \frac{ 3^xx ^{\sqrt{x }} }{ \left( x^4+2x \right)^{\frac{ 1 }{ 3 }} }\] \[=\ln \left( 3^xx ^{\sqrt{x}} \right)\ln \left( x^4+2x \right)^{\frac{ 1 }{ 3 }}\] \[=\ln 3^x+\ln x ^{\sqrt{x}}\frac{ 1 }{ 3 }\ln \left( x^4+2x \right)\] \[=x \ln 3+\sqrt{x}\ln x\frac{ 1 }{ 3 }\ln \left( x^4+2x \right)\] now you can differentiate.

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0Ohhh thanks so much @esamalla @surjithayer I get it now!!
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