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

1. FireKat97

2. anonymous

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

3. FireKat97

@esamalaa okay, what would be the next step?

4. anonymous

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

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

6. FireKat97

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

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

8. FireKat97

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

that is a very disgusting function haha

10. FireKat97

11. anonymous

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

12. FireKat97

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

13. anonymous

yw