## monroe17 3 years ago can someone help me step by step? find the derivative y=ln((e^(7x))/(sqrt(4x-5)))

1. satellite73

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

2. satellite73

it is the log of the whole thing right?

3. monroe17

yes.

4. satellite73

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

5. satellite73

are those steps clear?

6. monroe17

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

7. satellite73

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

8. satellite73

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

9. monroe17

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

10. Dido525

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

11. satellite73

the one half comes right out front as a multiplier

12. satellite73

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

13. satellite73

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

14. monroe17

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

15. satellite73

oh no not at all

16. satellite73

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

17. monroe17

okay

18. Dido525

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

19. satellite73

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. monroe17

gotcha. the power rule^

21. satellite73

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. satellite73

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

23. monroe17

okay imma try it/

24. satellite73

ok let me know what you get

25. Dido525

|dw:1351998238123:dw|

26. Dido525

It will be useful when you do the ln part.

27. monroe17

(1/4x-5)*4?

28. Dido525

Good!

29. monroe17

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

30. sirm3d

oops, you forgot to differentiate 7x

31. satellite73

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

32. satellite73

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

33. monroe17

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

34. satellite73

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. monroe17

woah im lost now.

36. monroe17

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

37. satellite73

yea it is right what you wrote is correct

38. satellite73

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

39. monroe17

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

40. satellite73

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

41. satellite73

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

42. monroe17

oh i get it now!

43. satellite73

the 7x should be 7

44. satellite73

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

45. monroe17

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

46. satellite73

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

47. satellite73

yup you got it

48. monroe17

thanks! im just a little slow sometimes lol