How to differentiate absolute value of multivariable function?

In summary: Remember the double angle formula for sine: $\sin(2x)=2 \sin(x) \cos(x)$. Therefore, $\sin(x) \cos(x)= \frac{ \sin(2x)}{2}$. And since $x+y$ is just another variable, $\sin(x+y) \cos(x+y)= \frac{ \sin(2(x+y))}{2}$. In summary, we can use the double angle formula for sine to simplify the expression $|\sin(x+y)|$ when taking its derivative with respect to $x$.
  • #1
Petrus
702
0
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?
\(\displaystyle f_x(x,y)=\frac{\sin(2x)}{2\sqrt{\sin^2(x+y)}}\)\(\displaystyle |\pi\rangle\)
 
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  • #2
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.
 
  • #3
Hey Petrus,

Check the argument of the sine function in the numerator...
 
  • #4
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,
\(\displaystyle |\pi\rangle\)
 
  • #5
Where did $y$ go in the numerator?
 
  • #6
MarkFL said:
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
\(\displaystyle \sin(x+y)=\sin(x) \cos(y)+ \cos(x) \sin(y)\)
and \(\displaystyle \sin^2(x)=\frac{1+ \cos(2x)}{2}\)
so I am back to spot one... How shall I do this..?

Regards,
\(\displaystyle |\pi\rangle\)
 
  • #7
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?
 
  • #8
Ackbach said:
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
\(\displaystyle |\pi\rangle\)
 
  • #9
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?
 
  • #10
Petrus said:
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?
 
  • #11
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 a lot better!

Regards,
\(\displaystyle |\pi\rangle\)
 
  • #12
Petrus said:
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 a lot better!

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

You got it.
 

FAQ: How to differentiate absolute value of multivariable function?

What does it mean to differentiate multivariable functions?

Differentiating multivariable functions means finding the rate of change of a function with respect to multiple variables. It involves finding the partial derivatives of the function with respect to each variable.

How is differentiating multivariable functions different from single variable functions?

In single variable functions, we only have one independent variable and the derivative is just the slope of the tangent line. In multivariable functions, we have multiple independent variables and the derivative is a vector that represents the direction and rate of change in each variable.

What is the chain rule in multivariable differentiation?

The chain rule in multivariable differentiation is used to find the derivative of a composition of functions. It states that the derivative of a composition of functions is equal to the product of the derivative of the outer function and the derivative of the inner function.

Can we differentiate a multivariable function with respect to more than one variable at a time?

Yes, we can differentiate a multivariable function with respect to more than one variable at a time. This is called a partial derivative, which measures the rate of change of the function with respect to a specific variable while holding all other variables constant.

How is the gradient related to multivariable differentiation?

The gradient is a vector that contains the partial derivatives of a multivariable function. It points in the direction of the steepest increase of the function and its magnitude represents the rate of change in that direction.

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