Animated Solution for Mathematics - Differentiation: The derivative of ${\sec }^{-1}\left(\frac{1}{2x^2-1}\right)$ with respect to $\sqrt{1-x^2}$ at $x=\frac{1}{2}$ is $\dots \dots \dots$

Welcome, future engineer! Today, we are going to dismantle a problem that, at first glance, looks like a tangled mess of inverse trigonometry. We are asked to find the derivative of $u={\sec }^{-1}\left(\frac{1}{2x^2-1}\right)$ with respect to $v=\sqrt{1-x^2}$ at the point $x=\frac{1}{2}$.

It looks intimidating, but the secret to mastering JEE calculus is not brute force—it is recognizing the hidden structure.

The Transformation

Whenever you see a function defined in terms of another, the most powerful tool in your arsenal is parametric differentiation. We define our two functions:

Our goal is to find $\frac{du}{dv}$. The chain rule tells us that:

Before we dive into differentiation, let us simplify $u$. That inverse secant is a trap. Recall the identity ${\sec }^{-1}\left(\frac{1}{\theta }\right)={\cos }^{-1}(\theta )$. By applying this, our function $u$ instantly transforms into $u={\cos }^{-1}(2x^2-1)$.

The Trigonometric Key

Now, we could differentiate ${\cos }^{-1}(2x^2-1)$ directly, but why make life hard? Let us use a clever substitution. If we set $x=\cos \theta$, then $2x^2-1$ becomes $2{\cos }^2\theta -1$.

Does that look familiar? It is the classic double-angle identity for cosine:

So, our function becomes $u={\cos }^{-1}(\cos (2\theta ))$. Since the inverse cosine and cosine functions are inverses of each other, they effectively neutralize, leaving us with $u=2\theta$. Substituting back $\theta ={\cos }^{-1}x$, we get the incredibly simple expression:

The Parametric Dance

Now that we have $u=2{\cos }^{-1}x$, finding $\frac{du}{dx}$ is a breeze. The derivative of ${\cos }^{-1}x$ is $\frac{-1}{\sqrt{1-x^2}}$, so:

Next, we tackle $v=\sqrt{1-x^2}$. Using the chain rule:

The derivative of $1-x^2$ is $-2x$, so:

The Grand Finale

We have our two derivatives. Now, we perform the final division:

Look at the beauty of this cancellation! The square root terms vanish, the negative signs vanish, and we are left with:

Finally, we evaluate this at $x=\frac{1}{2}$. Plugging it in, we get:

And there you have it! By simplifying the trigonometry first, we turned a terrifying problem into a simple arithmetic step. Keep this mindset, and you will conquer any calculus problem the JEE throws at you. The final answer is 4.