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

1. satellite73 Group Title

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

2. satellite73 Group Title

it is the log of the whole thing right?

3. monroe17 Group Title

yes.

4. satellite73 Group Title

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

5. satellite73 Group Title

are those steps clear?

6. monroe17 Group Title

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

7. satellite73 Group Title

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

8. satellite73 Group Title

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

9. monroe17 Group Title

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

10. Dido525 Group Title

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

11. satellite73 Group Title

the one half comes right out front as a multiplier

12. satellite73 Group Title

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

13. satellite73 Group Title

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

14. monroe17 Group Title

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

15. satellite73 Group Title

oh no not at all

16. satellite73 Group Title

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

17. monroe17 Group Title

okay

18. Dido525 Group Title

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

19. satellite73 Group Title

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 Group Title

gotcha. the power rule^

21. satellite73 Group Title

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 Group Title

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

23. monroe17 Group Title

okay imma try it/

24. satellite73 Group Title

ok let me know what you get

25. Dido525 Group Title

|dw:1351998238123:dw|

26. Dido525 Group Title

It will be useful when you do the ln part.

27. monroe17 Group Title

(1/4x-5)*4?

28. Dido525 Group Title

Good!

29. monroe17 Group Title

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

30. sirm3d Group Title

oops, you forgot to differentiate 7x

31. satellite73 Group Title

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

32. satellite73 Group Title

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

33. monroe17 Group Title

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

34. satellite73 Group Title

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 Group Title

woah im lost now.

36. monroe17 Group Title

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

37. satellite73 Group Title

yea it is right what you wrote is correct

38. satellite73 Group Title

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

39. monroe17 Group Title

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

40. satellite73 Group Title

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 Group Title

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

42. monroe17 Group Title

oh i get it now!

43. satellite73 Group Title

the 7x should be 7

44. satellite73 Group Title

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

45. monroe17 Group Title

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

46. satellite73 Group Title

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

47. satellite73 Group Title

yup you got it

48. monroe17 Group Title

thanks! im just a little slow sometimes lol