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1)The functions $f$ and $g$: $\mathbb{R} \rightarrow \mathbb{R} $ shall be 3-times differentiable.
Calculate $(f \cdot g)^{(3)}$.

1) $(f \cdot g)'=(f'g+fg')$

$(f'g+fg')'= (f''g+f'g')+(f'g'+fg'')= f''g+2f'g'+fg''$

$(f''g+2f'g'+fg'')'=(f'''g+f''g')+2(f''g'+f'g'')+(f'g''+fg''')$

$=f'''g+3(f''g'+f'g'')+fg'''=(f \cdot g)^{(3)}$

2)Find a function f:$\mathbb{R} \rightarrow \mathbb{R} $, which is 2-times differentiable on $\mathbb{R}$

2)$f(x)=x^2$
$f'(x)=2x$ and $f''(x)=2$

Are my solutions correct or did I sth. wrong?

Phil
  • 549

2 Answers2

1

Yes, indeed, your solutions are correct. Nice work.

Did you note the pattern that terms and the coefficients of $(fg), (fg)', (fg)'', (fg)'''$ follow?

Does Pascal's triangle look familiar?:

enter image description here

See Binomial Theorem and note the correlation.

amWhy
  • 209,954
1

Your work looks fine. Now try to prove that $$(f\cdot g)^{(n)}=\sum_{k=0}^n \binom{n}{k}f^{(k)}g^{(n-k)}$$

and that

$(1)$ Any polynomial function is $m$ times differentiable, for any $m$ you pick.

$(2)$ The function $$x^{m+1}\sin \left(\frac 1 x\right)$$ is $m$ times differentiable at $x=0$, but not $m+1$ times differentiable at $x=0$.

Pedro
  • 122,002