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: Recall 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 \(y = f(x)\).

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

Given \(f(x)=\ln(2x + e)\). Using the chain - rule \((\ln(u))^\prime=\frac{u^\prime}{u}\), where \(u = 2x+e\) and \(u^\prime=2\). So \(f^\prime(x)=\frac{2}{2x + e}\).

Step3: Determine the value of \(x\) corresponding to the given point \((y,x)=(1,0)\)

We know that when \(y = 1\) (the \(y\) - value of the point \((1,0)\)), \(x = 0\) (from the point \((1,0)\)).

Step4: Evaluate \(f^\prime(x)\) at \(x = 0\)

Substitute \(x = 0\) into \(f^\prime(x)\): \(f^\prime(0)=\frac{2}{2\times0 + e}=\frac{2}{e}\).

Step5: Use the formula for \((f^{-1})^\prime(y)\)

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

Answer:

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