Find derivative of Taylor series

Question

How would one differentiate a function in this form?

$$f(x) = f(a)+{\frac {f'(a)}{1!}}(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+{\frac {f'''(a)}{3!}}(x-a)^{3}$$

I'm sorry if this is something obvious, I'm not great with maths. I tried searching for the answer with no luck.

Answer 1

Differentiate with respect to $x$ ? On the left you have simply $f'(x)$ and on the right you have polynomial in $x$ ($a$ is a constant with respect to the derivative in $x$), so you get $$0+f'(a)+f''(a)(x-a)+\frac{f'''(a)}{2}(x-a)^2$$.

Answer 2

Each term has the form $$t_n(x) = \frac{(x-a)^n}{n!} \lim_{x \rightarrow a} \left( \frac{{\rm d}^n}{{\rm d}x^n} f(x) \right)$$

The derivative of such term is

$$t_n'(x) = \left( \frac{{\rm d} }{{\rm d}x} \frac{ (x-a)^n}{n!} \right) \lim_{x \rightarrow a} \left( \frac{{\rm d}^n}{{\rm d}x^n} f(x) \right)$$

with the result

$$ t_n'(x) = \frac{(x-a)^{n-1}}{(n-1)!} \lim_{x \rightarrow a} \left( \frac{{\rm d}^n}{{\rm d}x^n} f(x) \right) $$

In the end $f'(x) \approx \sum \limits_{n=1}^\infty t_n'(x) $

Answer 3

The derivative is really simple: it is identical to the original, but the first term is deleted (because it is a constant), and all the other coefficients are shifted to the left (except for the factorial denominator)

$$f(x) = f(a)+{\frac {\color{Blue}{f'(a)}}{1!}}(x-a)+{ \frac {\color{Red}{ f''(a)}}{2!}}(x-a)^{2}+{\frac {\color{Magenta}{f'''(a)}}{3!}}(x-a)^{3} \\ \\ f'(x) ={\color{Blue}{f'(a)}}+ { \frac {\color{Red}{f''(a)}}{1!}}(x-a)+{\frac {\color{Magenta}{f'''(a)}}{2!}}(x-a)^{2}+{\frac {f''''(a)}{3!}}(x-a)^{3}$$

You can check it, by deriving each term of the summation.