## anonymous one year ago derivative of y=x^(3/2)log(base3)squareroot(x+7)

1. zepdrix

$$\large\rm y=x^{3/2}\cdot\log_3\sqrt{x+7}$$ This guy? Hmm this looks like a fun one :) lol

2. anonymous

$y= x ^{3/2}\log_{3}\sqrt{x+7}$

3. anonymous

yeah lol

4. zepdrix

Do you remember your change of base formula for logs?$\large\rm \log_{a}(b)=\frac{\ln(b)}{\ln(a)}$We would choose natural log like this ^ Because it's easier to differentiate.

5. zepdrix

Unless you have a special formula memorized for log derivatives :) $\large\rm y=x^{3/2}\cdot\frac{\ln\sqrt{x+7}}{\ln3}$

6. anonymous

so it would be $(3/2)x ^{1/2}*(\ln \sqrt{x+7}/\ln3)$

7. anonymous

?

8. zepdrix

No no no :O Product rule.

9. anonymous

what should i do then?

10. zepdrix

I'll do the "set up" and you can let me know if it makes sense:$\large\rm y'=\color{royalblue}{\left(x^{3/2}\right)'}\cdot\frac{\ln\sqrt{x+7}}{\ln3}+x^{3/2}\cdot\color{royalblue}{\left(\frac{\ln\sqrt{x+7}}{\ln3}\right)'}$This is our product rule "set up". I haven't taken any derivatives yet. I'm just setting it up. We need to differentiate the blue parts.

11. anonymous

is the first blue part $(3/2)x ^{(3/2)-1}$

12. zepdrix

$\large\rm y'=\color{orangered}{\left(\frac{3}{2}x^{1/2}\right)}\cdot\frac{\ln\sqrt{x+7}}{\ln3}+x^{3/2}\cdot\color{royalblue}{\left(\frac{\ln\sqrt{x+7}}{\ln3}\right)'}$Ok ya that sounds right!

13. zepdrix

This other guy is a bit tricky though, ya? :)

14. anonymous

$1/((1/2)x ^{-1/2}+(1/2)7^{-1/2}) /(1/3)$

15. anonymous

?

16. zepdrix

No. And oh boy, did you ever make things complicated :D lol

17. zepdrix

Let's back up a sec, apply a helpful log rule before we take the derivative.

18. anonymous

sorry lol

19. anonymous

$\ln \sqrt{x+7}-\ln 3$

20. anonymous

?

21. zepdrix

First notice that ln3 is just a constant, so we can pull it out of the differentiation process and ignore it.$\large\rm \color{royalblue}{\left(\frac{\ln\sqrt{x+7}}{\ln3}\right)'}=\frac{1}{\ln3}\color{royalblue}{\left(\ln\sqrt{x+7}\right)'}$

22. anonymous

23. zepdrix

Next rewrite your square root as a 1/2 power,$\large\rm \frac{1}{\ln3}\color{royalblue}{\left(\ln(x+7)^{1/2}\right)'}$And apply another log rule,$\large\rm =\frac{1}{\ln3}\color{royalblue}{\left(\frac{1}{2}\ln(x+7)\right)'}$And again pull the 1/2 out front because it's just a constant,$\large\rm =\frac{1}{2\ln3}\color{royalblue}{\left(\ln(x+7)\right)'}$The problem becomes a lot easier to deal with at this point.

24. zepdrix

This is one of the things that can make Calc a difficult subject, remember all of these silly Algebra rules :) They're really helpful though.

25. anonymous

does the x+7 become 8

26. anonymous

or does it become 2

27. zepdrix

I don't understand what you're asking

28. anonymous

does $(\ln(x+7))\prime = \ln(2)\prime = 1/2$

29. zepdrix

?? 0_o

30. zepdrix

I have no idea where that 2 is coming from.. this is so confusing

31. zepdrix

Do you remember your derivative for ln(x) ?

32. anonymous

isnt the derivative of a constant = 1?

33. anonymous

so the derivative of x = 1 and the derivative of 7 = 1

34. anonymous

so 1+1 = 2

35. anonymous

???

36. zepdrix

The derivative of a constant is = 0. x is not a constant. But let's not worry about that. That's not what we're dealing with here. I'm asking if you know the derivative of $$\large\rm \ln(x)$$

37. anonymous

is it 1?

38. zepdrix

... No :(

39. anonymous

isnt the derivative of ln(x) = 1/x

40. zepdrix

Yes.

41. zepdrix

So then by that same rule,$\large\rm \frac{d}{dx}\ln(x+7)=\frac{1}{(x+7)}$

42. anonymous

wait so is the answer $(3/2)x ^{1/2}-((\ln(\sqrt{x+7})/\ln3)*x ^{3/2})*(1/x+7)$

43. zepdrix

No :(

44. anonymous

45. zepdrix

By the steps I was showing you before, the derivative of that second blue thing we were working on, would be,$\large\rm \color{royalblue}{\left(\frac{\ln\sqrt{x+7}}{\ln3}\right)'}=\color{orangered}{\frac{1}{2\ln3(x+7)}}$

46. zepdrix

$\large\rm y'=\color{orangered}{\left(\frac{3}{2}x^{1/2}\right)}\cdot\frac{\ln\sqrt{x+7}}{\ln3}+x^{3/2}\cdot\color{royalblue}{\left(\frac{\ln\sqrt{x+7}}{\ln3}\right)'}$ So I guess our final result is something like this,$\large\rm y'=\color{orangered}{\left(\frac{3}{2}x^{1/2}\right)}\cdot\frac{\ln\sqrt{x+7}}{\ln3}+x^{3/2}\cdot\color{orangered}{\frac{1}{(2\ln3)(x+7)}}$

47. zepdrix

Practice more problems cp :O Really really need to study!

48. anonymous

OH OKAY

49. anonymous

I GET IT NOW

50. anonymous

THANK YOU SO MUCH