## anonymous 3 years ago can someone help me step by step? find the derivative y=ln((e^(7x))/(sqrt(4x-5)))

1. anonymous

before you begin taking the derivative, use the properties of the log to make this expression easier to differentiate

2. anonymous

it is the log of the whole thing right?

3. anonymous

yes.

4. anonymous

so first step would be, before beginning to take the derivative, rewrite as $\ln(e^{7x})-\frac{1}{2}\ln(4x-5)$

5. anonymous

are those steps clear?

6. anonymous

why is 1/2 there and not (4x-5)^(1/2)

7. anonymous

i used two facts $\log(\frac{a}{b})=\log(a)-\log(b)$ and $\log(a^n)=n\log(a)$

8. anonymous

because $\log(\sqrt{4x-5})=\log((4x-5)^{\frac{1}{2}})=\frac{1}{2}\log(4x-5)$

9. anonymous

but 1/2 is outside the () and not inside ?

10. anonymous

You can use the rule like that. As long as it's insside the log.

11. anonymous

the one half comes right out front as a multiplier

12. anonymous

on other words, $$\log(\sqrt{x})=\frac{1}{2}\log(x)$$

13. anonymous

then one more step before differentiating since log and exp are inverse functions, you have $\log(e^{7x})=7x$

14. anonymous

okay, so do i need to use product rule for 1/2ln(4x-5)?

15. anonymous

oh no not at all

16. anonymous

$$\frac{1}{2}$$ is just a constant , leave it there

17. anonymous

okay

18. anonymous

Well you could... But it's a waste of time.

19. anonymous

for example if you wanted the derivative of $$\frac{1}{2}x^3$$ you do not use the product rule, you just say $$\frac{3}{2}x^2$$

20. anonymous

gotcha. the power rule^

21. anonymous

so now you have $7x+\frac{1}{2}\ln(4x-5)$ so the only rule you need now is the chain rule for the second part, and also knowing what the derivative of the log is

22. anonymous

typo there, i meant $7x-\frac{1}{2}\ln(4x-5)$

23. anonymous

okay imma try it/

24. anonymous

ok let me know what you get

25. anonymous

|dw:1351998238123:dw|

26. anonymous

It will be useful when you do the ln part.

27. anonymous

(1/4x-5)*4?

28. anonymous

Good!

29. anonymous

7x-1/2*(1/4x-5)*4

30. anonymous

oops, you forgot to differentiate 7x

31. anonymous

careful of the first term derivative of $$7x$$ is just $$7$$

32. anonymous

and you might not want to write $-\frac{4}{2(4x-5)}$ since you can cancel a 2

33. anonymous

i thought i was just doing chain rule? i know that that's 7?

34. anonymous

chain rule for $$-\frac{1}{2}\ln(4x-5)$$ because it is a composition , the log of something for $$7x$$ that is just a line with slope 7, derivative is 7

35. anonymous

woah im lost now.

36. anonymous

i did the chain rule for ln(4x-5) not 1/2*ln(4x-5)

37. anonymous

yea it is right what you wrote is correct

38. anonymous

the "minus one half" is just a constant, leave it there like you did

39. anonymous

okay, so where do i go from there. that fraction you did confused me.

40. anonymous

the only mistake in your answer was that you left $$7x$$ there, when the derivative of $$7x$$ is $$7$$ everything else was right

41. anonymous

you wrote this 7x-1/2*(1/4x-5)*4

42. anonymous

oh i get it now!

43. anonymous

the 7x should be 7

44. anonymous

and the four is in the numerator, cancels with the 2 in the denominator

45. anonymous

derivative of 7x is 7. 1/2 of 4 is 2. 7-2/4x-5

46. anonymous

so it should look more like $$7-\frac{2}{4x-5}$$

47. anonymous

yup you got it

48. anonymous

thanks! im just a little slow sometimes lol