Question

In 2A-12 Question A (http://goo.gl/o4IYp) I am getting \[1-2x-5x^2/2.\] for the quadratic approximation instead of Jerison's \[1+2x+5x^2/2.\]

As the derivative of
\[\frac{e^x}{1-x}\]

is

\[\frac{e^x(x-2)}{(x-1)^2}\]

and we are solving for x=0 I fail to see where my answer is wrong.

Answer 1

The derivative of \[\frac{e^x}{1-x}\] is \[\frac{e^x}{1-x} +\frac{e^x}{(1-x)^2} = \frac{e^x(2-x)}{(1-x)^2}\]

When I take derivatives like this, I use the product rule and the chain rule. So, I rewrite the equation first: \[f(x) = \frac{e^x}{1-x} = e^x(1-x)^{-1}\]


Hmmm, did you swap (1-x) for (x-1) in the denominator maybe?