[GRE 1268, #28a] Suppose that \(f\) is an invertible function, and the line \(y=3x+1\) is tangent to the graph of \(f(x)\) at \(x=0\). Let \(g(x)=\ln (1+x)\). Find the following:

1. \(f'(0)\)
2. \(\left(f^{-1}\right)' (1)\)
3. \((fg)'(0)\)
4. \( (f\circ g)'(0)\)
5. \((g\circ f)'(0)\)

Solution We are given a function, and its tangent line \(y=3x+1\) at a given point. The slope of the tangent line is the derivative at that point. So this tells us that \(f'(0)=3\), which answers the first question. It also tells us that \(f(0)=1\) (because the line \(y=3x+1\) goes through \((0,1)\)). To find information about the inverse function, write the equation that defines it: \[ f^{-1}(f(x)) = x.\] Take the derivative of both sides, using the Chain Rule: \begin{eqnarray*}\frac{d}{dx}\left(f^{-1}(f(x))\right) &=&\frac{d}{dx} x\\ \left(f^{-1}\right)'(f(x)) f'(x) &=&1 \end{eqnarray*} Now substitute \(x=0\): \begin{eqnarray*} \left(f^{-1}\right)'(f(0)) f'(0) &=&1\\ 3\left(f^{-1}\right)'(1) &=&1 \end{eqnarray*} So we conclude that \(\left(f^{-1}\right)'(1)=1/3\), answering the second question. To answer the third question, we need \(g(0)\) and \(g'(0)\). \begin{eqnarray*}g(x)&=&\ln (1+x)\\ g'(x)&=& \frac{1}{1+x}\\ g(0)&=&\ln(1)=0\\ g'(0)&=& \frac{1}{1}=1 \end{eqnarray*} So using the product rule we find \[(fg)'(0) = f'(0)g(0) + f(0)g'(0) = (3)(0) + (1)(1)=1.\] For the other two questions, \[(f\circ g)'(0)=f'(g(0))g'(0) = f'(0)g'(0)=(3)(1)=3,\] and \[(g\circ f)'(0)=g'(f(0))f'(0) = g'(1)f'(0)=\frac{1}{2} \cdot (3)=\frac{3}{2}.\]