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The question : Find $y^{(n)}(x)$ if $y(x)=\frac{1}{2-x}$... there is no explanation for $y^{(n)}$ in my textbook...can you explain this to me? First I tried to find the derivative of $\frac{1}{2-x}$ but then what?...

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Really? This isn't explained in your calculus textbook? How to find an n-th derivative? – Simon Hayward Nov 30 '12 at 11:00

As you wrote in your comment to @Florian's answer, you got some of the first derivatives $$ y'(x) = \frac{1}{(1-x)^2}, y''(x) = -\frac 2{(1-x)^3}, y'''(x) = \frac{2\cdot 3}{(1-x)^4}, \ldots $$ Now you have to identify a pattern in these derivatives ... some hints:

  • alternating signs, $+$ for odd orders (1,3,5,...), $-$ for even ones
  • in the numerator we have $2 \cdot 3 \cdot 4 \cdots$ upto what number? Can you link it to the order of the derivative?
  • in the denominator, we have $(1-x)^?$. How is the exponent linked to the order of the derivative.

Your pattern gives you a guess for $y^{(n)}(x) = \ldots$. Then prove you are right by induction (differentiating your guess).

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Are you sure you have the signs correct? – Mike Nov 30 '12 at 10:42

$y^{(n)}$ denotes the $n$-th derivative of $y$.

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I wrote the equation as y=(1-x)^-1 I get y'=(1-x)^2 ..and y" =-2* (1-x)^ do i write this in terms of y^n? – Imnothere Nov 30 '12 at 10:11

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