## DLS 2 years ago siny=xsin(a+y) dy/dx=?

1. DLS

I have a small doubt here..

2. DLS

is xsin(a+y) to be solved with product rule?

3. lgbasallote

dy/dx means you take the derivative of x... since sin(a+y) does not include any x in it... no, you don't use product rule. you treat sin(a+y) as a constant coefficient

4. DLS

:o?

5. hartnn

ummm.... u need to use product rule...

6. DLS

I did..

7. hartnn

because y is the function of x

8. DLS

What I want to ask is..if I had ysin(a+y) then I wuldve used chain rule?

9. lgbasallote

you only use product rule when two facots include x for example xsin x <-- x and sinx both have x so you use product rule xsiny <--- siny doesn't have x so treat it as coefficient remember that @hartnn ?

10. hartnn

no, product again, because both are functions of x

11. hartnn

sin y doesn't have x, right bu since y is the function of x....

12. hartnn

d/dx sin y = cos y dy/dx

13. DLS

ive done that only

14. DLS

but even WFA treats it as 0

15. hartnn

WFA ?

16. DLS

wolfaram :P ALPHA

17. DLS

spelling/*

18. hartnn

no...x sin(a+y) needs product rule.

19. DLS

:/

20. hartnn

x cos(a+y) dy/dx + sin(a+y)

21. hartnn

cos y dy/dx =x cos(a+y) dy/dx + sin(a+y) isolate dy/dx

22. hartnn
23. hartnn

it doesn't treat it as 0.

24. DLS

oh nvm :o

25. DLS

but still not clear why product rule..its used in product of 2 functions so x and sin(a+y) x is not a function,y is

26. hartnn

x is a function of x x^1

27. DLS

doesnt make any sense :S

28. DLS

every variable is afunction of itself then

29. hartnn

you can say that. f(x) = x^n with n=1.

30. DLS

why didnt we use chain rule for sin(a+y)

31. hartnn

we actually did.

32. DLS

:o

33. DLS

sin is one function and y other right?

34. hartnn

cos (a+y) d/dx(a+y) =cos (a+y) (0+dy/dx)

35. hartnn

yes.

36. DLS

so I have to treat x&y both as function not variables but other variable like abcd as constants(0) ?

37. DLS

when differentiating wrt x

38. RolyPoly

$\frac{d}{dx}xsin(a+y)$$=sin(a+y) \frac{d}{dx}(x) + x\frac{d}{dx}sin(a+y)$$=sin(a+y) + x[cos(a+y)\frac{d}{dx}(a+y)]$$=sin(a+y) + xcos(a+y)\frac{dy}{dx}$

39. hartnn

here since y is the function of x, we treat it as different function. if a is function of x, then a is also function..... if a is independent of x, then a is constant.

40. DLS

clear!

41. hartnn

:)