sin(cosx) differentiate (1 Viewer)

[Mellonie]

Y=sin(cosx)
what do u get when u differentiate that?
Y' = ?

(sori i know its easy question but it just not coming to me)

 

[mattchan]

Y=sin(cosx)

= -SinxCos(Cosx) ?

 

[Libbster]

This is a 'function of a function' rule q.

y= sin(cosx)
Remember to differentiate the 'main function' and then multiply by the derivative of function in the brackets
y'= cos(cosx) . -sinx
y' = -sinx.cos(cosx)

 

[Trev]

Through substitution:
y=sin(cosx)
let u = cosx
∴ y = sinu
du/dx = -sinx
dy/du = cosu
∴ dy/dx = du/dx*dy/du
= -sinx.cosu
dy/dx = -sinx.cos(cosx)

 

[wrong_turn]

i still think that libster's method is hte most accurate and the most in step with what you are meant to do:

ie. function of a fucntion rule:
y= x(angle)^u
y'= ux(angle)^u-1 . d/dx(angle)

 

[king_of_boredom]

as a general rule:

if y = sin f(x)
then y' = f'(x) cos f(x)

so if f(x) = cos x, therefore f'(x) = -sin x

so the answer is y' = -sin x.cos(cos x)

yeah everybody's gettin the same answer, must be right. funny thing is i've never seen a question like that before.

 

[acmilan]

i still think that libster's method is hte most accurate and the most in step with what you are meant to do:

ie. function of a fucntion rule:
y= x(angle)^u
y'= ux(angle)^u-1 . d/dx(angle)

er how is it more accurate than Trev's method? Trev essentially used the same method ie the chain rule dy/dx = dy/du*du/dx which is the definition of the chain rule. So if one is to be technical, Trev used a more formal approach to the question.

 

[acmilan]

as a general rule:

if y = sin f(x)
then y' = f'(x) cos f(x)

so if f(x) = cos x, therefore f'(x) = -sin x

so the answer is y' = -sin x.cos(cos x)

yeah everybody's gettin the same answer, must be right. funny thing is i've never seen a question like that before.

Ive seen much stranger ones then that.

eg.

sin(6cos(6sinx)) or sin(sin(sinx)) or ln(cos(1-t²))

 

[mattchan]

lol, on a side note

is the derivative to sin(6cos(6sinx))

= -36Cosx*Sin(6Sinx)*Cos(6Cos(6Sinx)) ?

 

[acmilan]

lol, on a side note

is the derivative to sin(6cos(6sinx))

= -36Cosx*Sin(6Sinx)*Cos(6Cos(6Sinx)) ?

that it is

 

[Trev]

The method I used was to allow more understanding for those who might not understand your method. I wouldn't bother with that much working out normally.

 

[Jago]

lol, on a side note

is the derivative to sin(6cos(6sinx))

= -36Cosx*Sin(6Sinx)*Cos(6Cos(6Sinx)) ?

Do you have to use the substitution method to get that answer?

 

[acmilan]

Do you have to use the substitution method to get that answer?

You just have to differentiate a few times. Remember that sin(u)' = u'cos(u)

In this case u = 6cos(6sinx)

[sin(6cos(6sinx))]'
= [(6cos(6sinx)]'cos(6cos(6sinx))
= [(6sinx)'*-6sin(6sinx)]cos(6cos(6sinx))
= [-36cosx.sin(6sinx)]cos(6cos(6sinx))
= -36cosx.sin(6sinx).cos(6cos(6sinx))

 

[Trev]

I think i'll mention a question like that to my 4 unit teacher in hope he puts one in our 3 unit paper

 

[Mellonie]

Howd u d it

lol, on a side note

is the derivative to sin(6cos(6sinx))

= -36Cosx*Sin(6Sinx)*Cos(6Cos(6Sinx)) ?

hey how doo u get that?

 

[jarrypan]

this one is right

Y=sin(cosx)

= -SinxCos(Cosx) ?

agree, it should be this answer

 

[acmilan]

hey how doo u get that?

I showed the working a few posts above

http://www.boredofstudies.org/community/showpost.php?p=1602045&postcount=13

 

[Riddles]

y=sin(cosx)
y'=-sinxcos(cosx)