## anonymous 3 years ago Find dy/dx for y=ln(5−x)^6

1. anonymous

$y = \ln(5-x)^6$ take derivative : $\frac{dy}{dx} = 6 \cdot \ln(5-x)^5 \frac{d(5-x)}{dx}$

2. anonymous

3. anonymous

It is by chain rule.. In chain rule we are using power rule also which says : $y = f(x)^n$ $\frac{dy}{dx} = n \cdot f(x)^{n-1} \cdot \frac{d(f(x))}{dx}$

4. anonymous

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

5. anonymous

$\frac{dy}{dx} = 6 \cdot \ln(5-x)^5 \frac{d(\ln(5-x)}{dx} \cdot \frac{d(5-x)}{dx}$

6. anonymous

ln((x-5)^6) = 6ln(x-5) ( 6ln(x-5) ) ' = 6/(x-5)

7. anonymous

Mixing up things...

8. anonymous

oh! duh! Got it, thank you!!1

9. anonymous

$y = \ln(5-x)^6$ $\frac{dy}{dx} = \frac{1}{(5-x)^6} \times 6 \times (5-x)^5 \times (-1)$

10. anonymous

Is this right???

11. anonymous

It will reduce to : $\frac{dy}{dx} = \frac{-1}{(5-x)}$

12. anonymous

ln((5-x)^6) = 6ln(5-x) ( 6ln(5-x) ) ' = -6/(5-x)

13. anonymous

Forgot 6 there.. Oh God...

14. anonymous

$\frac{dy}{dx} = \frac{-6}{(5-x)}$

15. anonymous

Yes by power rule, you can bring 6 in front: $y = 6 \cdot \ln(5-x)$ Now take the derivative..

16. anonymous

Okay, I've got it figured out! Thank you! If you'd like to help some more I just posted another question :)