Question

derivative of inverse function of y= log(2x-1)​

Answer 1

Let $y = f(x)= log_e(2x-1)$  

Let us find the inverse first i.e. $f^{-1} (x)$.

Step 1: Swap $x$ and $y$

$x = log_e(2y-1)$

Step 2: Find the value of $y$

Writing in exponential form:

$e^{x} = e^{log_e(2y-1)}\\\Rightarrow e^{x} = (2y-1)\\\Rightarrow y = \dfrac{1}{2}(e^{x} +1)$

Step 3: Write $y$ as $f^{-1} (x)$ i.e. inverse function.

$f^{-1} (x) =\dfrac{1}{2}(e^{x} +1)$

Now, taking derivative:

$\dfrac{d}{dx}(f^{-1}(x) ) = \dfrac{d}{dx}(\dfrac{e^{x} }{2}+\dfrac{1}{2} ) \\\Rightarrow \dfrac{1}{2} \dfrac{d}{dx}(e^{x}) + \dfrac{d}{dx} (\dfrac{1}{2})\\\Rightarrow \dfrac{e^{x} }{2}$

Formulas used above:

1. $\dfrac{d}{dx} (\text{ Constant}) = 0$

2. $\dfrac{d}{dx} (\text{ Constant }\times f(x)}) = \text{ Constant }\times \dfrac{d}{dx} (f(x)})$

3. $\dfrac{d}{dx} (e^{x} ) = e^{x}$

So, answer is $\dfrac{e^{x} }{2}$.

Answer 2

Step-by-step explanation:

y=log(2x-1) solve this following equation