# Fourth degree Maclaurin $f(x) = \dfrac{\sin x}{1-x}$

$$f(x) = \frac{\sin x}{1-x}$$

I tried to differentiate it but it is very difficult, what is the trick?

• You could find the terms up to order $4$ of the product of the series for $\sin x$ and $1/(1-x)$. – David Mitra Jul 30 '13 at 0:20
• @DavidMitra Why not make that an answer? – Pedro Tamaroff Jul 30 '13 at 0:23
• If you decide to make it an answer, @David, leave a comment on mine, and I'll delete it. – Cameron Buie Jul 30 '13 at 0:24
• See here for a related problem – Mhenni Benghorbal Jul 30 '13 at 10:13

I agree with David. Find the series for $\sin(x)$ and $\dfrac1{1-x}$ (a much simpler task), then find the first several terms of the product.
• @PaulthePirate $\sin x = x-x^3/3!+\cdots$ and $1/(1-x)=1+x+x^2+x^3+\cdots$. Just find the terms up to order $4$ of $( x-x^3/3!+\cdots)(1+x+x^2+x^3+\cdots )$. Take care to note that in the product the "dots" give terms $c_k x^k$ with $k>4$, so you can ignore them. – David Mitra Jul 30 '13 at 0:33
• @PaulthePirate: As tomasz points out, the product of the two Maclaurin series for $\sin(x)$ and $\dfrac1{1-x}$ will be precisely the Maclaurin series for $f(x)$. David has been gracious enough to narrow your parameters further, but you should be certain that you understand why you don't actually need the whole Maclaurin series for $\sin(x)$ or $\dfrac1{1-x},$ so that you can draw conclusions appropriately for future problems. – Cameron Buie Jul 30 '13 at 0:40