Question

consider the following function. without finding the inverse, evaluate the derivative of the inverse at the given point.
$f(x)=ln (2 x+e):(1,0)$
the derivative of the inverse at the given point is
(type an exact answer.)

Explanation:

Step1: Apply the formula for the derivative of the inverse function

If \(y = f(x)\) and \(x = f^{-1}(y)\), then \((f^{-1})^\prime(y)=\frac{1}{f^\prime(x)}\), where \(y = f(x)\). Given the point \((1,0)\), when \(y = 0\), \(x = 1\) (since if \(f(x)=y\), for the inverse function \(f^{-1}(y)=x\)).

Step2: Find the derivative of \(f(x)\)

First, find the derivative of \(f(x)=\ln(2x + e)\). Using the chain - rule, if \(u = 2x+e\), then \(f(x)=\ln(u)\) and \(\frac{df}{dx}=\frac{d}{du}(\ln(u))\cdot\frac{du}{dx}\).
We know that \(\frac{d}{du}(\ln(u))=\frac{1}{u}\) and \(\frac{du}{dx}=2\). So \(f^\prime(x)=\frac{2}{2x + e}\).

Step3: Evaluate \(f^\prime(x)\) at \(x = 1\)

Substitute \(x = 1\) into \(f^\prime(x)\). Then \(f^\prime(1)=\frac{2}{2\times1+e}=\frac{2}{2 + e}\).

Step4: Use the formula for \((f^{-1})^\prime(y)\)

Since \((f^{-1})^\prime(y)=\frac{1}{f^\prime(x)}\) and \(y = 0,x = 1\), \((f^{-1})^\prime(0)=\frac{1}{f^\prime(1)}\).
Substituting \(f^\prime(1)=\frac{2}{2 + e}\), we get \((f^{-1})^\prime(0)=\frac{2 + e}{2}\).

Answer:

\(\frac{e + 2}{2}\)