# ²/t²...

#### dr.power

Joined Oct 11, 2011
4
Hello guys,

I am newbie here, and this id my first post.
Somebody suggested me to ask my question here.

Sorry but I have some problems calculating the the derivatives and integrals of sine waves in forms of sin^2 wt.

I have a circuit which it's input has the form of "Vmsinwt" I added a dc offset to it as Vdc, So the final input has the forms of (Vdc + Vmsinwt).

The output of my circuit when the input signal is as (Vdc + Vmsinwt) would be in forms of (Vc + Vmsinwt) where Vc is the peak amplitude of a sine wave too which of course is just equal to Vdc.
Now I want to square the said ouput and then take double derivative of it.
By squaring, we'll get this: (Vc + Vmsinwt)². But I can not go on to take this:
δ²/δt² (Vc + Vmsinwt)² .
I just can take the derivative for sinx and cosx, Not for Vmsin² wt.
So plz kindly can you help me out HOW to do it? what is the method of doing so and whats the result plz?
What if the input happens to be Vmcoswt?

Thanks a bunch

Last edited:

#### Georacer

Joined Nov 25, 2009
5,181
Is this homework?

Why do you want to calculate the second derivative?

In general the following is true:
$$\left(A \cdot f^n(t) \right)'=A \cdot n \cdot f^{n-1}(t) \cdot f'(t)$$

#### panic mode

Joined Oct 10, 2011
1,783
you are differentiating with respect to t. everything else is considered constant so first derivative is:

δ/δt (Vc + Vmsinwt)²=
2 (Vc + Vmsinwt)*Vm*cos(wt)*w

then second derivative is:
δ/δt [2 (Vc + Vmsinwt)*Vm*cos(wt)*w]=
δ/δt [2w*Vm*Vc*cos(wt)+w*Vm²*2*sinwt*coswt]=
δ/δt [2w*Vm*Vc*cos(wt)+w*Vm²*sin(2*wt)]=
2*w²*Vm*Vc*(-sin(wt))+w²*Vm²*cos(2wt2)*2

#### panic mode

Joined Oct 10, 2011
1,783
lookup 'chain rule'.
it helps differentiating more complex terms like

sin²(wt)

δ/δt sin²(wt)
=2*sin(wt) * δ/δt sin(wt)
=2*sin(wt) * cos(wt)*δ/δt (wt)
=2*sin(wt) * cos(wt)*w
=sin(2wt)*w

you can also lookup wolfram alpha and bookmark it...
you will need it

#### dr.power

Joined Oct 11, 2011
4
Hi Panic,

you are differentiating with respect to t. everything else is considered constant so first derivative is:

δ/δt (Vc + Vmsinwt)²=
2 (Vc + Vmsinwt)*Vm*cos(wt)*w
Please can you let me know How did you calculate the above?

then second derivative is:
δ/δt [2 (Vc + Vmsinwt)*Vm*cos(wt)*w]=
δ/δt [2w*Vm*Vc*cos(wt)+w*Vm²*2*sinwt*coswt]=
δ/δt [2w*Vm*Vc*cos(wt)+w*Vm²*sin(2*wt)]=
2*w²*Vm*Vc*(-sin(wt))+w²*Vm²*cos(2wt2)*2
Can you Explain this too plz?
Are you sure regarding to the final response? Do'nt you think that it needs a thirs component?

#### Georacer

Joined Nov 25, 2009
5,181
The thread has been moved to the Math section, since the OP doesn't respond whether this is homework or not.

#### Ron H

Joined Apr 14, 2005
7,014
The thread has been moved to the Math section, since the OP doesn't respond whether this is homework or not.
He's been corresponding with me by PM. I encouraged him to post his questions on the forum, for the benefit of him and the members.
From what he said, this is not homework. It is a small part of a very ambitious project, which I will leave to him to disclose if he so wishes.

#### dr.power

Joined Oct 11, 2011
4
The thread has been moved to the Math section, since the OP doesn't respond whether this is homework or not.
Actually my question is not a homework,
I just want to know the response of my question because The output of my circuit has such a square then doble derivative response, And I want to know how it does look like.

Yet I am wondering why by derivation of (Vc + Vmsinwt)² we will reach to this!:
δ/δt (Vc + Vmsinwt)²=
2 (Vc + Vmsinwt)*Vm*cos(wt)*w

Can you guys plz help me out and enlighten me.

Joined Jul 7, 2009
1,583
It has already been explained to you that it's an application of the chain rule. This is taught in every beginning calculus class, but it sounds like you haven't taken calculus.

The basic principle is that when you have a function of a function (sometimes called function composition), the derivative of the composite function can be calculated using the chain rule.

$$\frac{d}{dx}f(u(x)) = \frac{df}{du} \cdot \frac{du}{dx}$$

This is what the people above used to do the calculation. I'm sure it looks mysterious when you first see it, but after working through some examples in a calculus text it becomes second nature. I'll refer you to a calculus text or the web. One good place to learn about it is here: http://www.khanacademy.org/#browse. Go to the calculus section and look at the chain rule videos.

#### dr.power

Joined Oct 11, 2011
4
It has already been explained to you that it's an application of the chain rule. This is taught in every beginning calculus class, but it sounds like you haven't taken calculus.

The basic principle is that when you have a function of a function (sometimes called function composition), the derivative of the composite function can be calculated using the chain rule.

$$\frac{d}{dx}f(u(x)) = \frac{df}{du} \cdot \frac{du}{dx}$$

This is what the people above used to do the calculation. I'm sure it looks mysterious when you first see it, but after working through some examples in a calculus text it becomes second nature. I'll refer you to a calculus text or the web. One good place to learn about it is here: http://www.khanacademy.org/#browse. Go to the calculus section and look at the chain rule videos.
Thanks I already knew about chain rule, But I have a small scale experience regarding to that. Thats why I couldnt do it my self.
Anyway I guess that this is not true:
δ/δt (Vc + Vmsinwt)²=
2 (Vc + Vmsinwt)*Vm*cos(wt)*w
Are you agreed with me?

Joined Jul 7, 2009
1,583
If Vc, Vm, and ω are constants, then the derivative given is correct. In other words

$$\frac{d}{dt}(V_c + V_m sin \omega t)^2 = [ 2\omega V_m cos \omega t] (V_c + V_m sin \omega t)$$

#### Georacer

Joined Nov 25, 2009
5,181
Okay, let's take it more slowly:
Keep in mind that $$\left( f \left( g(x) \right) \right) '=f'(x) \cdot \left(g(x) \right)'$$

$$\left( (V_c+V_m \sin(wt) )^2 \right)'= 2 (V_c +V_m \sin (wt) ) \cdot (V_c +V_m \sin (wt) )'= 2 (V_c +V_m \sin (wt) ) \cdot \left( (V_c)'+(V_m \sin(wt))' \right)= 2 (V_c +V_m \sin (wt) ) \cdot V_m \cdot \cos(wt) \cdot (wt)'= 2 (V_c +V_m \sin (wt) ) \cdot V_m \cdot \cos(wt) \cdot w$$

Is that clear?

#### magnet18

Joined Dec 22, 2010
1,227
Thanks I already knew about chain rule, But I have a small scale experience regarding to that. Thats why I couldnt do it my self.
Anyway I guess that this is not true:
δ/δt (Vc + Vmsinwt)²=
2 (Vc + Vmsinwt)*Vm*cos(wt)*w
Are you agreed with me?
Are you having trouble with the partial derivative part?

#### panic mode

Joined Oct 10, 2011
1,783
yep, he has trouble with differentiation...

here is link to interesting math site with plenty of videos explaining everything pretty well. there are several videos on chain rule:

http://patrickjmt.com/

hope that helps