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FireKat97

  • one year ago

Hi, how would you solve...

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  1. FireKat97
    • one year ago
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  2. anonymous
    • one year ago
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    first step: take ln for both sides and apply \[\ln\frac{ a }{ b }=lna-lnb\]

  3. FireKat97
    • one year ago
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    @esamalaa okay, what would be the next step?

  4. anonymous
    • one year ago
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    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 \[\ln(a)^{b}=b*lna\]

  5. anonymous
    • one year ago
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    all these steps are only for simplify the problem to be ready fir derivative

  6. FireKat97
    • one year ago
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    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)?

  7. anonymous
    • one year ago
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    no you have to differentiate the LHS too \[lny \rightarrow \frac{ 1 }{ y }*\frac{ dy }{ dx }\]

  8. FireKat97
    • one year ago
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    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 :)

  9. anonymous
    • one year ago
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    that is a very disgusting function haha

  10. FireKat97
    • one year ago
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    hahaha tell me about it

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

  12. FireKat97
    • one year ago
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    Ohhh thanks so much @esamalla @surjithayer I get it now!!

  13. anonymous
    • one year ago
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    yw

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