Question

If $f(x) = \tan^{-1}(\frac{\sqrt{1+x^2}-1}{x})$ then $f'(0)$ is:

• A. $\frac{1}{2}$
• B. 0
• C. 1
• D. non existant

Answer 1

Answer: (A)

1/2

Answer 2

To find $f'(0)$ for $f(x) = \tan^{-1}(\frac{\sqrt{1+x^2}-1}{x})$, we first simplify the expression for $f(x)$.

Step 1: Simplify the expression inside the $\tan^{-1}$ function.

Let $x = \tan\theta$. Since the domain of $\tan^{-1}x$ is $(-\infty, \infty)$, $\theta = \tan^{-1}x$ implies $\theta \in (-\pi/2, \pi/2)$.

Substitute $x = \tan\theta$ into the expression:

$$\frac{\sqrt{1+x^2}-1}{x} = \frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta}$$

We know that $1+\tan^2\theta = \sec^2\theta$.

$$= \frac{\sqrt{\sec^2\theta}-1}{\tan\theta}$$

Since $\theta \in (-\pi/2, \pi/2)$, $\cos\theta > 0$, which means $\sec\theta > 0$. Therefore, $\sqrt{\sec^2\theta} = |\sec\theta| = \sec\theta$.

$$= \frac{\sec\theta-1}{\tan\theta}$$

Convert $\sec\theta$ and $\tan\theta$ to $\sin\theta$ and $\cos\theta$:

$$= \frac{\frac{1}{\cos\theta}-1}{\frac{\sin\theta}{\cos\theta}} = \frac{\frac{1-\cos\theta}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}} = \frac{1-\cos\theta}{\sin\theta}$$

Now, use the half-angle trigonometric identities: $1-\cos\theta = 2\sin^2(\theta/2)$ and $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$.

$$= \frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)} = \frac{\sin(\theta/2)}{\cos(\theta/2)} = \tan(\theta/2)$$

Step 2: Determine the simplified form of $f(x)$.

Substitute the simplified expression back into $f(x)$:

$$f(x) = \tan^{-1}(\tan(\theta/2))$$

Since $\theta \in (-\pi/2, \pi/2)$, it follows that $\theta/2 \in (-\pi/4, \pi/4)$.

For any $y \in (-\pi/2, \pi/2)$, $\tan^{-1}(\tan y) = y$. Since $\theta/2 \in (-\pi/4, \pi/4)$, which is a subinterval of $(-\pi/2, \pi/2)$, we have:

$$f(x) = \theta/2$$

Now, substitute back $\theta = \tan^{-1}x$:

$$f(x) = \frac{1}{2}\tan^{-1}x$$

This simplified form is valid for all $x \in \mathbb{R}$.

For $x=0$, the original function's argument is $\frac{\sqrt{1+0^2}-1}{0} = \frac{0}{0}$, which is an indeterminate form.

We evaluate the limit: $\lim_{x \to 0} \frac{\sqrt{1+x^2}-1}{x} = \lim_{x \to 0} \frac{(\sqrt{1+x^2}-1)(\sqrt{1+x^2}+1)}{x(\sqrt{1+x^2}+1)} = \lim_{x \to 0} \frac{1+x^2-1}{x(\sqrt{1+x^2}+1)} = \lim_{x \to 0} \frac{x^2}{x(\sqrt{1+x^2}+1)} = \lim_{x \to 0} \frac{x}{\sqrt{1+x^2}+1} = \frac{0}{\sqrt{1}+1} = 0$.

So, $f(0) = \tan^{-1}(0) = 0$.

The simplified form $f(x) = \frac{1}{2}\tan^{-1}x$ also gives $f(0) = \frac{1}{2}\tan^{-1}(0) = 0$. Thus, $f(x) = \frac{1}{2}\tan^{-1}x$ is valid for all $x$.

Step 3: Calculate the derivative $f'(x)$.

Differentiate $f(x) = \frac{1}{2}\tan^{-1}x$ with respect to $x$:

$$f'(x) = \frac{d}{dx}\left(\frac{1}{2}\tan^{-1}x\right) = \frac{1}{2} \cdot \frac{1}{1+x^2}$$

Step 4: Evaluate $f'(0)$.

Substitute $x=0$ into $f'(x)$:

$$f'(0) = \frac{1}{2} \cdot \frac{1}{1+0^2} = \frac{1}{2} \cdot \frac{1}{1} = \frac{1}{2}$$