Originally Posted by

**xEnOn** I am trying to calculate the 11th derivative, $\displaystyle f^{(11)}(1)$, of a function $\displaystyle f(x) = \frac{x^{2}}{2+3x}$ using the Taylor Series centred at one.

I arrived to this stage where I form it into a geometry series format: $\displaystyle \frac{x^{2}}{5(1-(-\frac{3(x-1)}{5}))}$

From here, I believe that the $\displaystyle -\frac{3(x-1)}{5})$ is essentially the r in a geometry series formula.

So I put that into a summation: $\displaystyle \frac{x^{2}}{5}\sum_{n=0}^{\infty}(-\frac{3(x-1)}{5})$

I cleaned it up a little bit and manipulated the algebra and arrived at: $\displaystyle

\sum_{n=0}^{\infty}\frac{1}{5}(-\frac{3}{5})^{n}x^{2}(x-1)^{n}$

Now, this looks like a Taylor Series format to me with $\displaystyle (x-1)^{n}$ at the back and $\displaystyle \frac{1}{5}(-\frac{3}{5})^{n}x^{2}$ as my co-efficient.

Since I got the co-efficient of the series, I thought I could just do this: $\displaystyle \frac{f^{(11)}(1)}{11!} = \frac{1}{5}(-\frac{3}{5})^{n}x^{2}$

and then get: $\displaystyle f^{(11)}(1) = 11!\frac{1}{5}(-\frac{3}{5})^{n}x^{2}$

However, my answer is wrong and when I do it on a calculator, $\displaystyle \frac{\partial^{11} }{\partial x^{11}}(\frac{x^{2}}{2+3x}) \neq 11!\frac{1}{5}(-\frac{3}{5})^{n}x^{2}$

What should I do after this? I thought I can get the answer once I turn it into a format that looks like the original Taylor series format and find the answer from the co-efficient?

thanks!