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sin(cosx) differentiate (1 Viewer)

Mellonie

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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)
 

Libbster

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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

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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

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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)
 
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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

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wrong_turn said:
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

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king_of_boredom said:
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

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lol, on a side note

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

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

Jago

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mattchan said:
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

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Jago said:
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))
 

Mellonie

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Howd u d it

mattchan said:
lol, on a side note

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

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



hey how doo u get that?
 

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