Leibnitz Formulae for differentiation - Tricky question

[thomas49th]

All I know about Leibnitz is

$$\frac{d^{n}}{dx^{n}}f(x) = \sum^{n}_{i=0} \left(\stackrel{n}{i}\right) g^{n-i}(x)h^{i}(x)$$

and I don't understand what the answer sheet says:

Parts of it perhaps - I can see they take the 0th, 1st and 2nd derivative, but why not more? I mean why stop at 2? I know m>=2 for y^(m+2) to be +ve, but if m was 3 (which it could be) why do we not have the 3rd derivative in there aswell? Also using the combination for working out the binomal coefficients, taking the 1st derivative for example, how does $$\frac{(m-1)!}{(m-2)!1!} = m$$? I get m-1.

Stumped.

 

[HallsofIvy]

All I know about Leibnitz is

$$\frac{d^{n}}{dx^{n}}f(x) = \sum^{n}_{i=0} \left(\stackrel{n}{i}\right) g^{n-i}(x)h^{i}(x)$$

That doesn't even make sense- unless you say f(x)= g(x)h(x) first!

and I don't understand what the answer sheet says:

Parts of it perhaps - I can see they take the 0th, 1st and 2nd derivative, but why not more? I mean why stop at 2? I know m>=2 for y^(m+2) to be +ve, but if m was 3 (which it could be) why do we not have the 3rd derivative in there aswell? Also using the combination for working out the binomal coefficients, taking the 1st derivative for example, how does $$\frac{(m-1)!}{(m-2)!1!} = m$$? I get m-1.

Here, $f(x)= 4x^2+ 1$ so that $f'(x)= 8x$, $f''(x)= 8$ and $f'''(x)= 0$. It is the fact that the third derivative (and all higher) derivatives of f are 0 that is important, not derivatives of y. Indeed, there is no reason to think that y is a polynomial at all,

Stumped.

 

[thomas49th]

Here, $$f(x)= 4x^{2} + 1$$ so that , and . It is the fact that the third derivative (and all higher) derivatives of f are 0 that is important, not derivatives of y. Indeed, there is no reason to think that y is a polynomial at all,

How come f(x) equals the coefficient of y?