Given $f:R^n \rightarrow R$ and $x:R \rightarrow R^n$ such that $f$ possesses 2nd order parial derivative and $x(t)=a+th, \ a,h\in R^n$. I want to prove that $g=f(x(t))$ is twice differentiable and $g''(0)=(h.D)^2f(x(0))$.
$$Dg=Df(x(t)).Dx(t)=Df(x(t)).h$$
$$D^2g=D(Df(x(t)).h)=D(Df(x(t))).h $$
How to proceed furthur ?
I basically want to use this for proving Taylor's theorem for several variable so source for that could also be helpful.
Higher order derivatives of several variables.
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calculus
real-analysis
multivariable-calculus
derivatives
1 Answers
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As $ x'(t) = h$ and $x''(t) = 1$, $x(t)$ is twice diferentiable and as $f$ is twice diferentiable, $ g = f o x = f(x(t))$ is twice diferentiable.
From here: $$D^2g=D(Df(x(t)).h)$$ from the product property you have: $$ D(Df(x(t)).h)= D^2(f(x(t))). h + Df(x(t)).D(h)$$ You get: $$= D^2f(x(t)).Dx(t). h + Df(x(t)).0 = D^2f(x(t)).h^2 $$
So $$g''(t) = D^2f(x(t)).h^2 $$ And finally: $$g''(0) = h^2.D^2f(x(0))$$