Here's the question you clicked on:
anlnt4
Find dy/dx for y=ln(5−x)^6
\[y = \ln(5-x)^6\] take derivative : \[\frac{dy}{dx} = 6 \cdot \ln(5-x)^5 \frac{d(5-x)}{dx}\]
It is by chain rule.. In chain rule we are using power rule also which says : \[y = f(x)^n\] \[\frac{dy}{dx} = n \cdot f(x)^{n-1} \cdot \frac{d(f(x))}{dx}\]
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
\[\frac{dy}{dx} = 6 \cdot \ln(5-x)^5 \frac{d(\ln(5-x)}{dx} \cdot \frac{d(5-x)}{dx}\]
ln((x-5)^6) = 6ln(x-5) ( 6ln(x-5) ) ' = 6/(x-5)
Mixing up things...
oh! duh! Got it, thank you!!1
\[y = \ln(5-x)^6\] \[\frac{dy}{dx} = \frac{1}{(5-x)^6} \times 6 \times (5-x)^5 \times (-1)\]
It will reduce to : \[\frac{dy}{dx} = \frac{-1}{(5-x)}\]
ln((5-x)^6) = 6ln(5-x) ( 6ln(5-x) ) ' = -6/(5-x)
Forgot 6 there.. Oh God...
\[\frac{dy}{dx} = \frac{-6}{(5-x)}\]
Yes by power rule, you can bring 6 in front: \[y = 6 \cdot \ln(5-x)\] Now take the derivative..
Okay, I've got it figured out! Thank you! If you'd like to help some more I just posted another question :)