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f(x)= x/x+1, find f'(x) using defnition of the derivative
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f(x) = A/B ... let A = x and B = x+1 f'(x) = (AB' - A'B)/B^2 f'(x) = (x(1) - 1(x+1))/(x+1)^2 f'(x) = (x - x - 1)/(x+1)^2 f'(x) = -1/(x+1)^2
using the definition, \[f'(x) = \lim_{\epsilon \rightarrow 0} \frac{\frac{x+\epsilon}{x+\epsilon+1} -\frac{x}{x+1} }{\epsilon} = \lim_{\epsilon \rightarrow 0} \frac{\epsilon}{(x+1)^2 + \epsilon(x+1)} \frac{1}{\epsilon} = \frac{1}{(x+1)^2} \]
it's 1/(x+1)^2
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