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If $f(0) =0$ and $f''(x)$ exists then what is the value of $$f'(x)-frac{f(x)}{x}$$

I tried Tailor and Lagrange theorems but to no avail.

clearly $f''(c)=2frac{f(x)-xf'(0
)}{x^2}$ where $0<c<x$
but how can i proceed further.
How can i find $f'(0)$

Expanding $f(t)$ around $x$ gives $f(t)=f(x)+(t-x)f'(x)+frac{(t-x)^2}2f''(tau)$, where $tau$ lies between $t$ and $x$.

For $t=0$ this gives $0=f(0)=f(x)-xf'(x)+frac{x^2}2f''(theta x)$ with $theta in (0,1)$ and therefore $f'(x)-frac{f(x)}x=frac x2f''(theta x)$.

It is not clear what you are asking, though there are many functions that satisfy your conditions but yielding very different expressions.


For example, take $f(x)=x$ for which $f(0)=0, f''(x)$ exists and equals $0$.
That gives you an expression for $$f'(x)-frac{f(x)}{x}=1-1=0$$


and then take $g(x)=sinx$, $g(0)=0,g''(x)=-sinx$ for which

$$g'(x)-frac{g(x)}{x}=cosx-frac{sinx}{x}$$

Two different "values" for two functions that satisfy your conditions.

By Taylor,

$$f'(x)-frac{f(x)}x=frac x2f''(0)+r'(x)-frac{r(x)}x,$$ where $r(x)$ is the remainder (hopefully, the last two terms can be neglectible for small $x$).