Prove f(x)=||x||^2 is differentiable. Help plz?

**wxyj** · October 12th 2009, 04:19 PM

Can anyone help me with this question???

prove f(x)=||x||^2 is differentiable using the definition of differentiability
where x is a vector in R^n. and ||x|| is the length of the vecotr(Not allowed to use use components of x)
x=(x1,x2,...xn)

The definitoin is lim as h->0, [f(a+h)-f(a) -c*h]/h =0
where c is the gradient of f,the function is differentiable at point a

I have absolutely no idea where to start, it's not like you have two distinct vairables..x,y....and prove differentiable at a particular point, and I am not allowed to use the components of x.How do i prove it's differentiable everywhere...

Anyone give me some hints?

Thanks so much

**Jose27** · October 12th 2009, 05:25 PM

It's easy to see that $c*h = 2<x,h>$ ie. $2<x,h>=D_f(x)h$ use this and the fact that $\Vert x+h \Vert ^2 = ^\Vert x \Vert ^2 + 2<x,h> + \Vert h \Vert ^2$ and substitute in the definition: $\vert \frac{ f(x+h) - f(x) - D_f(x)h}{ \Vert h \Vert } \vert$ and see that this goes to zero as $\Vert h \Vert \rightarrow 0$ . ( $<,>$ denotes the interior product, I'm assuming you're using the usual norm in $\mathbb{R} ^n$ )

**wxyj** · October 12th 2009, 07:50 PM

Originally Posted by Jose27

It's easy to see that $c*h = <x,h>$ ie. $<x,h>=D_f(x)h$ use this and the fact that $\Vert x+h \Vert ^2 = ^\Vert x \Vert ^2 + 2<x,h> + \Vert h \Vert ^2$ and substitute in the definition: $\vert \frac{ f(x+h) - f(x) - D_f(x)h}{ \Vert h \Vert } \vert$ and see that this goes to zero as $\Vert h \Vert \rightarrow 0$ . ( $<,>$ denotes the interior product, I'm assuming you're using the usual norm in $\mathbb{R} ^n$ )

Thanks so much

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