Question

If $f(x) = tan^{-1}x + \ln x$ then find $\frac{dy}{dx} @ x = \frac{\pi}{4}$ where $y = f^{-1}(x)$.

Answer 1

$\frac{2}{3}$

Answer 2

The problem asks us to find the derivative of the inverse function $y = f^{-1}(x)$ at a specific point $x = \frac{\pi}{4}$, given the function $f(x) = \tan^{-1}x + \ln x$.

1. Understand the relationship between a function and its inverse: If $y = f^{-1}(x)$, it means $x = f(y)$.

2. Apply the inverse function differentiation rule: Differentiating $x = f(y)$ with respect to $x$ using the chain rule: $\frac{d}{dx}(x) = \frac{d}{dx}(f(y))$

   $1 = f'(y) \frac{dy}{dx}$

   Therefore, $\frac{dy}{dx} = \frac{1}{f'(y)}$.

3. Find the derivative of the original function $f(x)$: Given $f(x) = \tan^{-1}x + \ln x$. Differentiate $f(x)$ with respect to $x$: $f'(x) = \frac{d}{dx}(\tan^{-1}x) + \frac{d}{dx}(\ln x)$

   $f'(x) = \frac{1}{1+x^2} + \frac{1}{x}$

4. Find the corresponding value of $y$ for the given $x$ value: We need to find $\frac{dy}{dx}$ at $x = \frac{\pi}{4}$. This means we need to find the value of $y$ such that $f(y) = \frac{\pi}{4}$. Substitute $y$ into $f(x)$: $f(y) = \tan^{-1}y + \ln y$ Set $f(y) = \frac{\pi}{4}$: $\tan^{-1}y + \ln y = \frac{\pi}{4}$ By inspection, if we let $y=1$: $\tan^{-1}(1) + \ln(1) = \frac{\pi}{4} + 0 = \frac{\pi}{4}$. So, when $x = \frac{\pi}{4}$, the corresponding value of $y$ is $1$.

5. Evaluate $f'(y)$ at the found $y$ value: Substitute $y=1$ into the expression for $f'(y)$: $f'(1) = \frac{1}{1+1^2} + \frac{1}{1}$

   $f'(1) = \frac{1}{2} + 1$

   $f'(1) = \frac{3}{2}$

6. Calculate $\frac{dy}{dx}$: Using the inverse function differentiation rule: $\frac{dy}{dx} \bigg|_{x=\frac{\pi}{4}} = \frac{1}{f'(1)} = \frac{1}{\frac{3}{2}} = \frac{2}{3}$.