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Compute the derivatives of two given functions

Consider the functions

    \begin{align*}  f(x) &= \frac{1}{1 + \frac{1}{x}}, \qquad x \neq 0 \\  g(x) &= \frac{1}{1 + \frac{1}{f(x)}}. \end{align*}

Find thd derivatives f'(x) and g'(x).


First, we compute the derivative of f,

    \begin{align*}  f'(x) &= \frac{\frac{1}{x^2}}{\left( 1+ \frac{1}{x} \right)^2} \\  &= \frac{1}{(x+1)^2}, \end{align*}

where we multiplied the numerator and denominator by x^2 to get the second line.

Next, using the chain rule we know

    \[ \left( 1+ \frac{1}{f(x)} \right)' = \frac{-f'(x)}{(f(x))^2}. \]

So, then using the quotient rule and chain rule we can evaluate the derivative of g,

    \begin{align*}  g'(x) &= \frac{\frac{f'(x)}{(f(x))^2}}{\left(1+ \frac{1}{f(x)}\right)^2} \\  &= \frac{f'(x)}{(f(x))^2 \left( 1 + \frac{1}{f(x)} \right)^2} \\  &= \frac{f'(x)}{(f(x) + 1)^2}. \end{align*}

Then, plugging in the definition of f(x) and the expression for f'(x) which we already computed, we have,

    \begin{align*}  g'(x) &= \frac{ \frac{1}{(x+1)^2} }{\left( \frac{1}{1+ \frac{1}{x}} + 1 \right)^2} \\[9pt]  &= \frac{1}{(1+x)^2 \left( \frac{2x+1}{x+1} \right)^2} \\  &= \frac{1}{(2x+1)^2} \\  &= (2x+1)^{-2}.  \end{align*}

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