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

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

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

2. It's easy to see that $c*h = 2$ ie. $2=D_f(x)h$ use this and the fact that $\Vert x+h \Vert ^2 = ^\Vert x \Vert ^2 + 2 + \Vert h \Vert ^2$ 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 $\mathbb{R} ^n$)

3. Originally Posted by Jose27
It's easy to see that $c*h = $ ie. $=D_f(x)h$ use this and the fact that $\Vert x+h \Vert ^2 = ^\Vert x \Vert ^2 + 2 + \Vert h \Vert ^2$ 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