MHB How to differentiate absolute value of multivariable function?

[Petrus]

Hello MHB,
I am working with finding max and min value and I always hesitate when I got absolute value so the one I strugle is to derivate $$|\sin(x+y)|$$
if we do it respect to x is this correct?
$$f_x(x,y)=\frac{\sin(2x)}{2\sqrt{\sin^2(x+y)}}$$$$|\pi\rangle$$

 

[Ackbach]

Hmm. I think you're missing some stuff. One thing is certain:
$$ \frac{d}{dx} |x|= \frac{x}{|x|} = \frac{|x|}{x}= \text{signum}(x).$$
Using the chain rule, therefore, you'd have
$$ \frac{ \partial}{ \partial x}| \sin(x+y)|= \frac{ \sin(x+y)}{| \sin(x+y)|} \, \cos(x+y).$$
You could probably simplify this a bit using some trig identities, but I don't think it would come down to what you got. Also note that $|x|= \sqrt{x^{2}}$, which does agree with your denominator.

 

[MarkFL]

Hey Petrus,

Check the argument of the sine function in the numerator...

 

[Petrus]

Hello,
Thanks for the fast responed!:)
hmm.. I prob can't use the hanf angle formulas what I did and get it wrong, hm.. I don't see what to do the inner deferentiate, What do you mean Mark?

Regards,
$$|\pi\rangle$$

 

[MarkFL]

Where did $y$ go in the numerator?

 

[Petrus]

Where did $y$ go in the numerator?

Now I understand but I did not think clearly cause now when I think, I don't know what to do with numberator, we got two rules
$$\sin(x+y)=\sin(x) \cos(y)+ \cos(x) \sin(y)$$
and $$\sin^2(x)=\frac{1+ \cos(2x)}{2}$$
so I am back to spot one... How shall I do this..?

Regards,
$$|\pi\rangle$$

 

[Ackbach]

From $\sin(x+y) = \sin(x) \cos(y)+ \cos(x) \sin(y)$, set $x=y$ and obtain $\sin(2x)= \sin(x) \cos(x)+ \cos(x) \sin(x) = 2 \sin(x) \cos(x)$. Therefore,
$$\sin(x) \cos(x)= \frac{ \sin(2x)}{2}.$$
So what would $\sin(x+y) \cos(x+y)$ be?

 

[Petrus]

From $\sin(x+y) = \sin(x) \cos(y)+ \cos(x) \sin(y)$, set $x=y$ and obtain $\sin(2x)= \sin(x) \cos(x)+ \cos(x) \sin(x) = 2 \sin(x) \cos(x)$. Therefore,
$$\sin(x) \cos(x)= \frac{ \sin(2x)}{2}.$$
So what would $\sin(x+y) \cos(x+y)$ be?

Hello Ackbach,
$$\sin(x+y)=2 \sin(x) \cos(x)$$
$$\cos(x+y)= \cos(x) \cos(y)- \sin(y) \sin(x)$$ set $$x=y$$ and obtain
$$\cos(2x)= \cos^2(x)- \sin^2(x)$$
so $$\sin(x) \cos(x)= \frac{ \sin(2x)}{2}(\cos^2(x)- \sin^2(x)) $$

Regards
$$|\pi\rangle$$

 

[MarkFL]

In your original post, you are essentially positing:

$$2\sin(x+y)\cos(x+y)=\sin(2x)$$

Do you see the problem with this?

 

[Ackbach]

Hello Ackbach,
$$\sin(x+y)=2 \sin(x) \cos(x)$$
$$\cos(x+y)= \cos(x) \cos(y)- \sin(y) \sin(x)$$ set $$x=y$$ and obtain
$$\cos(2x)= \cos^2(x)- \sin^2(x)$$
so $$\sin(x) \cos(x)= \frac{ \sin(2x)}{2}(\cos^2(x)- \sin^2(x)) $$

Regards
$$|\pi\rangle$$

I don't think you're seeing the pattern here. What would $\sin(z) \cos(z)$ be? What would $ \sin( \xi) \cos( \xi)$ be? What would $ \sin( \aleph) \cos( \aleph)$ be? So what must $\sin(x+y) \cos(x+y)$ be?

 

[Petrus]

Hello,
I think I see now $$\sin(x)\cos(x)=\frac{\sin(2x)}{2} <=>\sin(x+y)\cos(x+y)=\frac{\sin(2(x+y))}{2} $$

Well it is really late and I am tired so I can't think clear right now, I need to sleep so hopefully tomorow I understand a lot better!

Regards,
$$|\pi\rangle$$

 

[Ackbach]

Hello,
I think I see now $$\sin(x)\cos(x)=\frac{\sin(2x)}{2} <=>\sin(x+y)\cos(x+y)=\frac{\sin(2(x+y))}{2} $$

Well it is really late and I am tired so I can't think clear right now, I need to sleep so hopefully tomorow I understand a lot better!

Regards,
$$|\pi\rangle$$

You got it.