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Differentiate y=-2sin3x ?

Hi, I got an answer of -2cos3x for this?
Also, if you differentiated say y=sin3x, is this cos3x ? Many thanks
no its -6cos3x
Reply 2
Avatar for Dizgurl
Dizgurl
15
You multiply the coefficient before the 'sin' by the number before the x.
Here, when the coefficient is 1:
y= sin 3x would differentiate to y= 3cos3x.

Can you try your example now?
Original post by Dizgurl
You multiply the coefficient before the 'sin' by the number before the x.
Here, when the coefficient is 1:
y= sin 3x would differentiate to y= 3cos3x.

Can you try your example now?

Thanks for the reply. That's where i went wrong i think in this question. So it would be -6cos3x for this example? Also, what rule does this use/how does this work. I used a really conveluted product rule to work this out and my answer was still wrong, but yours is a much simpler method.
Reply 4
Avatar for FhizZ
FhizZ
9
Basically sin(3x) is a compound function. Derivative of f(g(x)), by the chain rule, is f'(g(x))*g'(x) so in this case derivative of sin(3x) is cos(3x) * (3x)' which is cos(3x) * 3, so thats where the 3 is coming from.
Reply 5
Avatar for Dizgurl
Dizgurl
15
Original post by Bertybassett
Thanks for the reply. That's where i went wrong i think in this question. So it would be -6cos3x for this example? Also, what rule does this use/how does this work. I used a really conveluted product rule to work this out and my answer was still wrong, but yours is a much simpler method.

Yes, that is correct. :smile:

Say for example with sin 3x:
As another poster has mentioned, this is a composite function so you use the chain rule:
Here the g(x) is 3x, and the second layer would be sin g(x).
So I'll let 3x= u

u= 3x so du/dx= 3
y= sin u so dy/du= cos

In the chain rule you multiply both these values to remove the du, so you get just dy/dx:
=3cos u
Substitute u=3x back into the equation
= 3cos 3x

This principle also applies to your question too
(edited 5 years ago)
Original post by Dizgurl
Yes, that is correct. :smile:

Say for example with sin 3x:
As another poster has mentioned, this is a composite function so you use the chain rule:
Here the g(x) is 3x, and the second layer would be sin g(x).
So I'll let 3x= u

u= 3x so du/dx= 3
y= sin u so dy/du= cos

In the chain rule you multiply both these values to remove the du, so you get just dy/dx:
=3cos u
Substitute u=3x back into the equation
= 3cos 3x

This principle also applies to your question too


Thanks for the reply. If it was say y=-3cos2x , would the defferential of this then be 6sin2x ? How does this change with cos?
Reply 7
Avatar for Dizgurl
Dizgurl
15
Original post by Bertybassett
Thanks for the reply. If it was say y=-3cos2x , would the defferential of this then be 6sin2x ? How does this change with cos?

Yes, the differential will be 6sin 2x, because differentiating cos gives us -sin, so you multiply the coefficient by the number before the x and swap the sign at the end.
So there it would be:
- (-3 x 2) sin 2x
= 6sin2x

With the chain rule it would be exactly the same, g(x) would be 2x, and the f(x) would be cos g(x).
So you would do everything exactly the same as my previous example, except obviously we're now differentiating cos to -sin, instead of from sin to cos.

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