Let f be a real-valued function defined on the interval (-1, 1) such that e⁻ˣ f(x) = 2 + ∫₀ˣ √(t⁴ +

Q: Wait, why do we need to find $x$ such that $f(x) = 2$? Can't we just differentiate the inverse formula directly?

Great question! The derivative of an inverse function, $(f^{-1})'(y)$, is defined at a specific point $y$. To evaluate it, you must know the corresponding $x$ such that $f(x) = y$. Without finding that specific $x$, you cannot evaluate the derivative of the original function $f'(x)$.

Q: Why can't we just use energy conservation or some other physics trick here?

This is a pure calculus problem! While physics problems often use conservation laws, here we rely on the fundamental relationship between a function and its inverse. The 'conservation' here is the geometric symmetry across the line $y = x$.

Q: Is the Leibniz Rule really necessary? Can't I just integrate $\sqrt{t^4 + 1}$?

That integral is non-elementary! You cannot express it in terms of simple functions. The beauty of the Leibniz Rule is that it allows us to find the derivative of the integral without ever needing to evaluate the integral itself. It's a powerful shortcut!

Question

Let $f$ be a real-valued function defined on the interval (-1, 1) such that $e^{-x} f(x) = 2 + \int_{0}^{x} \sqrt{t^4 + 1} dt$, for all $x \in (-1, 1)$, and let $f^{-1}$ be the inverse function of $f$. Then $(f^{-1})'(2)$ is equal to

Options:

(A) 1

(B) 1/3

(C) 1/2

(D) 1/e

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the edge of a mathematical landscape, looking at the function $f(x)$ defined by the equation:

$$e^{-x} f(x) = 2 + \int_{0}^{x} \sqrt{t^4 + 1} dt$$

It looks intimidating, but in the world of JEE Advanced, intimidation is just a sign that you are about to learn something beautiful. Our goal is to find $(f^{-1})'(2)$.

### The Golden Rule of Inverses

Before we touch the integral, let's remember the fundamental connection between a function and its inverse. The derivative of an inverse function is the reciprocal of the derivative of the original function.

Specifically, the relationship is defined as:

$$(f^{-1})'(y) = \frac{1}{f'(x)}, 	ext{ where } y = f(x)$$

This means our entire mission is reduced to two simple steps: find the $x$ that gives $f(x) = 2$, and then find the slope $f'(x)$ at that point.

### The Search for the Magic Point

How do we find $x$ such that $f(x) = 2$? Look at the integral $\int_{0}^{x} \sqrt{t^4 + 1} dt$.

If we set $x = 0$, the upper and lower limits of the integral become identical, and the integral vanishes. Let's test $x = 0$ in our original equation:

$$e^{-0} f(0) = 2 + \int_{0}^{0} \sqrt{t^4 + 1} dt$$

Since $e^0 = 1$ and the integral is $0$, we get $1 \cdot f(0) = 2 + 0$, which means $f(0) = 2$. We have found our point! The function passes through $(0, 2)$, so the inverse function must pass through $(2, 0)$.

### The Calculus Heavy Lifting

Now, let's differentiate the entire equation with respect to $x$. On the left side, we have a product of two functions, $e^{-x}$ and $f(x)$, so we use the Product Rule:

$$\frac{d}{dx} [e^{-x} f(x)] = -e^{-x} f(x) + e^{-x} f'(x)$$

On the right side, we have the constant $2$ and an integral. The derivative of $2$ is $0$. For the integral, we use the Leibniz Rule, which tells us that the derivative of $\int_{0}^{x} g(t) dt$ is simply $g(x)$.

So, the derivative of $\int_{0}^{x} \sqrt{t^4 + 1} dt$ is $\sqrt{x^4 + 1}$. Putting it all together, we get the equation:

$$-e^{-x} f(x) + e^{-x} f'(x) = \sqrt{x^4 + 1}$$

### The Final Victory

We are almost there! We need $f'(0)$, so let's plug $x = 0$ into our new derivative equation:

$$-e^{0} f(0) + e^{0} f'(0) = \sqrt{0^4 + 1}$$

We already know $f(0) = 2$, so this becomes $-1 \cdot 2 + 1 \cdot f'(0) = 1$. Simplifying this gives $-2 + f'(0) = 1$, which leads us to $f'(0) = 3$.

Finally, using our inverse derivative formula:

$$(f^{-1})'(2) = \frac{1}{f'(0)} = \frac{1}{3}$$

And there it is! The complexity of the integral was just a mask, and through the power of the Leibniz Rule and the symmetry of inverses, we have arrived at the elegant solution of **$1/3$**.

Animated Solution for Mathematics - Differentiation: Let $f$ be a real-valued function defined on the interval (-1, 1) such that $e^{- x} f (x) = 2 + \int_{0}^{x} t^{4} + 1 d t$ , for all $x \in (- 1, 1)$ , and let $f^{- 1}$ be the inverse function of $f$ . Then $(f^{- 1})^{'} (2)$ is equal to

Select Answer:

Visualized Solution

Problem Objective

Inverse Derivative Formula

Finding $x$ for $y = 2$

Evaluating $f (0)$

The Inverse Point

Differentiating the Equation

Applying Product & Leibniz Rules

Substituting $x = 0$

Calculating $f^{'} (0)$

Final Inverse Derivative

The Sigma Insight: Inverse of a Function

Analyzing the Setup

The Golden Rule of Inverses

The Search for the Magic Point

The Calculus Heavy Lifting

The Final Victory

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Select Answer:

Visualized Solution

Problem Objective

Inverse Derivative Formula

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=2

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