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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.

pishpish
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  • Notice that the derivative of the Taylor development of $f$ is the Taylor development of $f'$ and it suffices to add a single quote everywhere. –  Mar 29 '17 at 13:38

3 Answers3

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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 $$0+f'(a)+f''(a)(x-a)+\frac{f'''(a)}{2}(x-a)^2$$.

Héhéhé
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  • Thanks, the (x - a) part was giving me troubles. Also, I think you made a typo where f'''(x) should be f'''(a). – pishpish Mar 29 '17 at 11:47
  • the derivative of $x-a$ is simply 1 because the derivative of a sum is the sum of the derivatives and the derivative of $x$ is 1 and the derivative of $a$ is 0. Yeah it should be $f'''(a)$ I fixed it. – Héhéhé Mar 29 '17 at 13:33
2

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.

Raxi Ral
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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) $

John Alexiou
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  • Why are you using limits instead of simply $f^{(n)}(a)$ ? – Héhéhé Mar 29 '17 at 14:25
  • I wanted to be clear that $f^n(a)$ has no dependence to $x$ it is the same thing as $x \rightarrow a$ in the end. I guess it is a habit that comes from programming. – John Alexiou Mar 29 '17 at 15:53