## bmelyk 3 years ago need someone to walk me through this question, please:

1. bmelyk

Use the appropriate rule or combination of rules to find the derivative of the function defined below. $y=2(\sin(6x)+\cos(x^3))^\frac{ 1 }{ 2 }$

2. bmelyk

not even sure how to start this one. ;( looks like it would the the chain rule of course.

3. hartnn

first what is derivative of x^{1/2} = ?

4. bmelyk

1/2x^-1/2

5. hartnn

we always go from outer function to inner function

6. hartnn

$$y'=2(1/2)(\sin(6x)+\cos(x^3))^\frac{ -1 }{ 2 }d/dx(\sin(6x)+\cos(x^3))$$

7. bmelyk

options for the right answer are: $a)2*\frac{ \cos(6x)-\sin(x^3) }{ (\sin(6x)+cox(x^3))^1/2 }$ $b)\frac{ \cos(6x) +\sin(x^3)}{ (\sin(6x)+\cos(x^3))^\frac{ 1 }{ 2 } }$ $c)\frac{ 12 }{ 6x^\frac{ 1 }{ 2 } }*\cos(6x)-\frac{ 6 }{ \cos(x^3)^\frac{ 1 }{ 2 } }*\sin(x^3)*x^2$ $d)\frac{ 1 }{ (\sin(6x)^\frac{ 1 }{ 2 }+\cos(x^3))^\frac{ 1 }{ 2 } }$ $e)\frac{ 6*\cos(6x)-3x^2*\sin(x^3) }{ (\sin(6x)+\cos(x^3))^\frac{ 1 }{ 2 } }$ f) same as e only 2 outside of the fraction. $g) \frac{ 2 }{ \sin(6x)+\cos(x^3))^1/2 }$ $h) \frac{ 6 }{ \sin(6x)^\frac{ 1 }{ 2 } }*\cos6x-\frac{ 3 }{ \cos(x^3)^\frac{ 1 }{ 2 } }*\sin(x^3)*x^2$

8. hartnn

u got my last comment ? how i got that ?

9. bmelyk

and i had that so far.

10. bmelyk

yesi got that.

11. hartnn

so what is d/dx(sin(6x)+cos(x^3)) ?

12. bmelyk

do i use the product rule for sin*6x and cos*x^3?

13. hartnn

there is no product, so NO u can treat them individually also what is d/dx (sin 6x) = ?

14. bmelyk

cos6x

15. bmelyk

-sinx^3

16. hartnn

only ?

17. hartnn

i mean only cos 6x ?

18. bmelyk

times 6?

19. hartnn

u need chain rule here too, right ?

20. hartnn

yes! times 6, because u'll have d/dx(6x)

21. bmelyk

cox6x*6

22. bmelyk

cos6x*6

23. bmelyk

????

24. hartnn

yup, what about 2nd term ? d/dx (cos (x^3)) = ?

25. bmelyk

-sinx^3(3x^2)

26. hartnn

absolutely correct! u got it :)

27. hartnn

so which option is it ?

28. bmelyk

e??

29. hartnn

put everything together, and u'll get it.

30. hartnn

yes, correct :)

31. bmelyk

awesome!

32. hartnn

i hope chain rule is clearer to u now.....