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waterineyesBest ResponseYou've already chosen the best response.0
\[y = \ln(5x)^6\] take derivative : \[\frac{dy}{dx} = 6 \cdot \ln(5x)^5 \frac{d(5x)}{dx}\]
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
It is by chain rule.. In chain rule we are using power rule also which says : \[y = f(x)^n\] \[\frac{dy}{dx} = n \cdot f(x)^{n1} \cdot \frac{d(f(x))}{dx}\]
 one year ago

anlnt4Best ResponseYou've already chosen the best response.0
Wait, isn't that in the case that the 6th power would be to the entire thing? I thought you brought the 6 in front since lna^b=blna
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
\[\frac{dy}{dx} = 6 \cdot \ln(5x)^5 \frac{d(\ln(5x)}{dx} \cdot \frac{d(5x)}{dx}\]
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
ln((x5)^6) = 6ln(x5) ( 6ln(x5) ) ' = 6/(x5)
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
Mixing up things...
 one year ago

anlnt4Best ResponseYou've already chosen the best response.0
oh! duh! Got it, thank you!!1
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
\[y = \ln(5x)^6\] \[\frac{dy}{dx} = \frac{1}{(5x)^6} \times 6 \times (5x)^5 \times (1)\]
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
It will reduce to : \[\frac{dy}{dx} = \frac{1}{(5x)}\]
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
ln((5x)^6) = 6ln(5x) ( 6ln(5x) ) ' = 6/(5x)
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
Forgot 6 there.. Oh God...
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
\[\frac{dy}{dx} = \frac{6}{(5x)}\]
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
Yes by power rule, you can bring 6 in front: \[y = 6 \cdot \ln(5x)\] Now take the derivative..
 one year ago

anlnt4Best ResponseYou've already chosen the best response.0
Okay, I've got it figured out! Thank you! If you'd like to help some more I just posted another question :)
 one year ago
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