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Differentiate multivariable

Petrus

Well-known member
Feb 21, 2013
739
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 \(\displaystyle |\sin(x+y)|\)
if we do it respect to x is this correct?
\(\displaystyle f_x(x,y)=\frac{\sin(2x)}{2\sqrt{\sin^2(x+y)}}\)

Best regards,
\(\displaystyle |\pi\rangle\)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
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

Administrator
Staff member
Feb 24, 2012
13,775
Hey Petrus,

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

Petrus

Well-known member
Feb 21, 2013
739
Hello,
Thanks for the fast responed!:)
hmm.. I prob cant 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,
\(\displaystyle |\pi\rangle\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Where did $y$ go in the numerator?
 

Petrus

Well-known member
Feb 21, 2013
739
Where did $y$ go in the numerator?
Now I understand but I did not think clearly cause now when I think, I dont know what to do with numberator, we got two rules
\(\displaystyle \sin(x+y)=\sin(x) \cos(y)+ \cos(x) \sin(y)\)
and \(\displaystyle \sin^2(x)=\frac{1+ \cos(2x)}{2}\)
so I am back to spot one... How shall I do this..?

Regards,
\(\displaystyle |\pi\rangle\)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
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

Well-known member
Feb 21, 2013
739
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,
\(\displaystyle \sin(x+y)=2 \sin(x) \cos(x)\)
\(\displaystyle \cos(x+y)= \cos(x) \cos(y)- \sin(y) \sin(x)\) set \(\displaystyle x=y\) and obtain
\(\displaystyle \cos(2x)= \cos^2(x)- \sin^2(x)\)
so \(\displaystyle \sin(x) \cos(x)= \frac{ \sin(2x)}{2}(\cos^2(x)- \sin^2(x)) \)

Regards
\(\displaystyle |\pi\rangle\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
In your original post, you are essentially positing:

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

Do you see the problem with this?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
Hello Ackbach,
\(\displaystyle \sin(x+y)=2 \sin(x) \cos(x)\)
\(\displaystyle \cos(x+y)= \cos(x) \cos(y)- \sin(y) \sin(x)\) set \(\displaystyle x=y\) and obtain
\(\displaystyle \cos(2x)= \cos^2(x)- \sin^2(x)\)
so \(\displaystyle \sin(x) \cos(x)= \frac{ \sin(2x)}{2}(\cos^2(x)- \sin^2(x)) \)

Regards
\(\displaystyle |\pi\rangle\)
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

Well-known member
Feb 21, 2013
739
Hello,
I think I see now \(\displaystyle \sin(x)\cos(x)=\frac{\sin(2x)}{2} <=>\sin(x+y)\cos(x+y)=\frac{\sin(2(x+y))}{2} \)

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 alot better!

Regards,
\(\displaystyle |\pi\rangle\)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
Hello,
I think I see now \(\displaystyle \sin(x)\cos(x)=\frac{\sin(2x)}{2} <=>\sin(x+y)\cos(x+y)=\frac{\sin(2(x+y))}{2} \)

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 alot better!

Regards,
\(\displaystyle |\pi\rangle\)
You got it.