MHB Hence, the derivative of d/dx (x*x') is (d/dx)x*x'+x*x

  • Thread starter Thread starter Fernando Revilla
  • Start date Start date
  • Tags Tags
    Derivative
AI Thread Summary
The discussion focuses on finding the derivative of the expression d/dx (x*x'), where x is a vector and x' is its transpose, resulting in a matrix rather than a norm. The derivative of matrix products is established using the product rule, which states that the derivative of the product of two matrices A and B is given by the sum of the derivative of A multiplied by B and A multiplied by the derivative of B. Applying this rule to the specific case of AA^T leads to the conclusion that the derivative can be expressed as the sum of two terms involving the derivatives of A and its transpose. This mathematical framework is crucial for understanding derivatives in the context of matrix calculus. The discussion provides clarity on the differentiation of matrix products, emphasizing the importance of the product rule in this context.
Fernando Revilla
Gold Member
MHB
Messages
631
Reaction score
0
I quote a question from Yahoo! Answers

What is the derivative of d/dx (x*x') where x is a vector and x' denotes x transpose (note that x*x' is a matrix, and not the norm of x!)

I have given a link to the topic there so the OP can see my response.
 
Mathematics news on Phys.org
In general, if $A=[a_{ij}(x)],\;B=[b_{ij}(x)]$ are matrices $n\times n$ with $a_{ij}(x)$ and $b_{ij}(x)$ differentiable real funtions defined on the open interval $(\alpha,\beta)$, its derivative matrices are defined by $$\frac{d}{dx}A=\left[\frac{d}{dx}a_{ij}(x)\right],\;\frac{d}{dx}B=\left[\frac{d}{dx}b_{ij}(x)\right]$$ Using the product formula: $$A(x)B(x)=C(x)=[c_{ij}(x)] \quad \left(c_{ij}(x)=\sum_{k=1}^na_{ik}(x)b_{kj}(x) \right)$$
it is easy to prove the relation on $(\alpha,\beta)$ $$\frac{d}{dx}(AB)=\left(\frac{d}{dx}A\right)B+A \left(\frac{d}{dx}B\right) $$
as a consequence $$\frac{d}{dx}(AA^T)=\left(\frac{d}{dx}A\right)A^T+A\left(\frac{d}{dx}A^T\right)=\left(\frac{d}{dx}A\right)A^T+A\left(\frac{d}{dx}A\right)^T$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Back
Top