## anonymous one year ago Can someone please help me find the derivative of this problem? I will fan and medal.

1. anonymous

Whats the function?

2. anonymous

I'm typing it

3. anonymous

$\frac{ 2 }{ 3 }\sin^\frac{ 3 }{ 2 }x-\frac{ 2 }{ 7 }\sin^\frac{ 7 }{ 2 }x$

4. anonymous

$\frac{2}{3}\left(\sin^{2/3}x\right)' \iff \frac{2}{3}\left((\sin(x))^{2/3}\right)'$

5. wmj259

If you look carefully you see that the two functions are connected with the subtract sign. So you can break that function into 2 separate derivatives.

6. wmj259

$\frac{ d }{ dx } (\frac{ 2 }{ 3 }\sin ^{2/3}(x)) - \frac{ d }{ dx }((\frac{ 2 }{ 7 }\sin ^{7/2}x))$

7. anonymous

Use the power rule, then the chain rule.

8. wmj259

|dw:1444350991307:dw|

9. anonymous

lmao -_-

10. anonymous

Wait. How did you guys turn the exponent 3/2 into 2/3?

11. anonymous

Typo.

12. wmj259

yes that was a typo on my part, I apologize.

13. anonymous

Y'all are fine. I was just making sure.

14. anonymous

$\large \frac{d}{dx}\left[\frac{2}{3}(\sin(x))^{3/2}\right] = \frac{2}{3} \cdot \frac{3}{2} \cdot (\sin(x))^{(3/2) - 1} \cdot \frac{d}{dx} (\sin(x))$

15. wmj259

What I would do in this situation would be to first split the problem into 2 different differentiation's because of that subtract sign. You can pull out the (2/3) and the (2/7) because they are constants And then I would do the Power Rule on the (3/2) and the (7/2). Separately. And then I would do the chain rule on the 2 sin(x) functions.

16. wmj259

|dw:1444351292444:dw|

17. anonymous

$\large \frac{d}{dx}\left[\frac{2}{7}(\sin(x))^{7/2}\right] = \frac{2}{7} \cdot \frac{7}{2} \cdot (\sin(x))^{(7/2)-1} \cdot \frac{d}{dx}(\sin(x))$

18. wmj259

|dw:1444351320870:dw|

19. wmj259

|dw:1444351344236:dw|

20. anonymous

$\sin^\frac{ 1 }{ 2 }x*cosx-\sin^\frac{ 5 }{ 2 }x*cosx$

21. anonymous

That's what I got

22. anonymous

For which part?

23. anonymous

The entire problem

24. anonymous

Alright let's check

25. anonymous

I just did the power rule then chain rule as recommended.

26. wmj259

|dw:1444351512241:dw|

27. wmj259

Is the finished derivative for the left half.

28. wmj259

the simplified left side is |dw:1444351605530:dw|

29. wmj259

Now the right side would give us: |dw:1444351641222:dw|

30. wmj259

The simplified right side would be |dw:1444351722466:dw|

31. wmj259

So if you combine the left and the right side you get: |dw:1444351765027:dw|

32. wmj259

@Catseyeglint911, Which is what you got.

33. anonymous

@wmj259 Can you simplify the right side any further?

34. wmj259

you can say this: |dw:1444351917329:dw|

35. wmj259

|dw:1444351988066:dw|

36. anonymous

\begin{align}\large \frac{d}{dx}\left[\frac{2}{3}(\sin(x))^{3/2}\right] &= \frac{2}{3} \cdot \frac{3}{2} \cdot (\sin(x))^{(3/2) - 1} \cdot \frac{d}{dx} (\sin(x)) \\&= \sin^{1/2}(x)\cos(x) \ \end{align}\begin{align} \large \frac{d}{dx}\left[\frac{2}{7}(\sin(x))^{7/2}\right] &= \frac{2}{7} \cdot \frac{7}{2} \cdot (\sin(x))^{(7/2)-1} \cdot \frac{d}{dx}(\sin(x)) \\&=\sin^{5/2}(x)\cos(x)\end{align}\begin{align}\frac{d}{dx}\left[ \frac{ 2 }{ 3 }\sin^\frac{ 3 }{ 2 }x-\frac{ 2 }{ 7 }\sin^\frac{ 7 }{ 2 }x\right] &= \sin^{1/2}(x)\cos(x) -\sin^{5/2}(x)\cos(x) \\&= \sin^{1/2}(x)\cos(x) \left[ 1-\sin^{5}(x) \right] \end{align}

37. anonymous

$\large \color{red}{\boxed{\sin^{1/2}(x)\cos(x)\left[1-\sin^{5}(x)\right]}}$

38. anonymous

@wmj259 you did something funky here|dw:1444352122644:dw|

39. wmj259

Yes, The right most part should have been the derivative of sin(x) not just (x).

40. wmj259

It goes much longer then that because their is basically a Power rule then 2 chain rules. but the 2nd chain rule is always ignore because its just the derivative of x's in some cases. But with practice comes great speed.

41. anonymous

Thank you both! I wish I could give out multiple medals. I'll just become a fan of both of you. :)

42. anonymous

Woohoo!