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anonymous
 4 years ago
Find dy/dx for y=ln(5−x)^6
anonymous
 4 years ago
Find dy/dx for y=ln(5−x)^6

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wait, 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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ln((x5)^6) = 6ln(x5) ( 6ln(x5) ) ' = 6/(x5)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh! duh! Got it, thank you!!1

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ln((5x)^6) = 6ln(5x) ( 6ln(5x) ) ' = 6/(5x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Forgot 6 there.. Oh God...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{dy}{dx} = \frac{6}{(5x)}\]

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

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