Question

If $y = \text{Sec}^{-1}x$, then $\frac{d^2y}{dx^2} =$

• A. $\frac{1-2x^2}{x|x|(x^2-1)^{3/2}}$
• B. $\frac{1-x^2}{x^2(x^2-1)^{3/2}}$
• C. $-\frac{1-x^2}{x^2(x^2-1)^{3/2}}$
• D. $\frac{1+2x^2}{x|x|(x^2-1)^{3/2}}$

Hint:

When differentiating expressions involving $|x|$, it is often safest to consider the cases $x>0$ and $x<0$ separately. After finding the derivative for one case (e.g., $x>0$), you can often generalize the result by replacing terms like $x^2$ with $x|x|$ to make it valid for both cases.

Answer 1

The Correct Option is A

We are given $y = \text{Sec}^{-1}x$.
The first derivative is a standard result: 
$\frac{dy}{dx} = \frac{1}{|x|\sqrt{x^2-1}}$. 
To find the second derivative, it's easier to consider the cases for $|x|$. Let's assume $x>1$, so $|x|=x$. 
$\frac{dy}{dx} = \frac{1}{x\sqrt{x^2-1}} = (x(x^2-1)^{1/2})^{-1}$. 
Now, differentiate this expression with respect to $x$ using the chain rule and the product rule. 
$\frac{d^2y}{dx^2} = -1 \cdot (x(x^2-1)^{1/2})^{-2} \cdot \frac{d}{dx}(x\sqrt{x^2-1})$. 
$\frac{d^2y}{dx^2} = -\frac{1}{x^2(x^2-1)} \cdot \left[ 1 \cdot \sqrt{x^2-1} + x \cdot \frac{1}{2\sqrt{x^2-1}} \cdot 2x \right]$. 
$\frac{d^2y}{dx^2} = -\frac{1}{x^2(x^2-1)} \cdot \left[ \sqrt{x^2-1} + \frac{x^2}{\sqrt{x^2-1}} \right]$. 
Combine the terms inside the bracket by finding a common denominator: 
$\frac{d^2y}{dx^2} = -\frac{1}{x^2(x^2-1)} \cdot \left[ \frac{(x^2-1) + x^2}{\sqrt{x^2-1}} \right]$. 
$\frac{d^2y}{dx^2} = -\frac{1}{x^2(x^2-1)} \cdot \frac{2x^2-1}{(x^2-1)^{1/2}} = -\frac{2x^2-1}{x^2(x^2-1)^{3/2}}$. 
$\frac{d^2y}{dx^2} = \frac{1-2x^2}{x^2(x^2-1)^{3/2}}$. 
For $x>1$, we have $x^2 = x \cdot x = x|x|$. So we can write: 
$\frac{d^2y}{dx^2} = \frac{1-2x^2}{x|x|(x^2-1)^{3/2}}$. 
A similar calculation for $x<-1$ yields the same final expression.