If y = f((2x-1)/(x² + 1)) and f'(x) = sin x², then dy/dx = .........

Q: Why can't I just integrate $f'(x) = \sin(x^2)$ to find $f(x)$ first?

That is a brilliant thought, but it is a trap! The function $\sin(x^2)$ does not have an elementary antiderivative. You are not meant to find $f(x)$; you are meant to use the properties of the derivative directly. Trust the Chain Rule!

Q: Do I need to simplify the inner function $\frac{2x-1}{x^2+1}$ before differentiating?

Not at all! Simplifying it might actually make it more complex. The Quotient Rule is designed to handle rational functions exactly as they are. Just set up your $u$ and $v$ carefully and proceed.

Q: Why is the negative sign in the numerator so important?

It is the most common place to lose marks in JEE. When you apply the Quotient Rule, you have a minus sign between the two terms in the numerator. If you don't distribute that negative sign correctly across the entire second term, your final answer will be wrong. Always use parentheses!

Question

If $y = f\left(\frac{2x-1}{x^2+1}\right)$ and $f'(x) = \sin x^2$, then $\frac{dy}{dx} = \dots\dots\dots$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

In the world of JEE calculus, a composite function like $y = f\left(\frac{2x-1}{x^2+1}\right)$ represents a multi-layered puzzle. It consists of an outer shell, $f$, and an inner core, $g(x) = \frac{2x-1}{x^2+1}$.

When we calculate $\frac{dy}{dx}$, we are measuring how the entire system changes as $x$ varies. We do not need the explicit form of $f(x)$; we only require its rate of change, $f'(x) = \sin(x^2)$, and the rate of change of the inner core.

### Phase 1: The Chain Rule Strategy

The Chain Rule is the relay race of calculus. It states that the derivative of a composite function is the derivative of the outer function, evaluated at the inner function, multiplied by the derivative of the inner function itself.

Mathematically, this is expressed as:

$$\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$$

First, we handle the outer shell. Given $f'(x) = \sin(x^2)$, the term $f'(g(x))$ becomes:

$$f'(g(x)) = \sin\left(\left(\frac{2x-1}{x^2+1}\right)^2\right)$$

### Phase 2: The Quotient Rule Dance

Now we apply the Quotient Rule to the inner core, $g'(x) = \frac{d}{dx}\left(\frac{2x-1}{x^2+1}\right)$. To differentiate a fraction $\frac{u}{v}$, we use the formula:

$$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2}$$

Let $u = 2x-1$ and $v = x^2+1$. Consequently, $u' = 2$ and $v' = 2x$.

Assembling these pieces, we get:

$$\frac{(x^2+1)(2) - (2x-1)(2x)}{(x^2+1)^2}$$

Expanding the numerator carefully:

$$(2x^2 + 2) - (4x^2 - 2x) = 2x^2 + 2 - 4x^2 + 2x = -2x^2 + 2x + 2$$

Thus, the derivative of the inner core is:

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

### Phase 3: The Synthesis

We now combine the outer derivative and the inner derivative to reach the final result. Multiplying the components together, we obtain:

$$\frac{dy}{dx} = \left[ \sin\left(\left(\frac{2x-1}{x^2+1}\right)^2\right) \right] \cdot \left[ \frac{-2x^2 + 2x + 2}{(x^2+1)^2} \right]$$

The final simplified expression is:

$$\mathbf{\frac{dy}{dx} = \frac{2+2x-2x^2}{(x^2+1)^2} \sin\left(\left(\frac{2x-1}{x^2+1}\right)^2\right)}$$

This result demonstrates the essence of JEE Advanced mathematics: not brute force, but the systematic application of fundamental principles. Every complex problem is simply a series of manageable steps.

Animated Solution for Mathematics - Differentiation: If $y = f (\frac{2 x - 1}{x ^{2} + 1})$ and $f^{'} (x) = sin x^{2}$ , then $\frac{d y}{d x} = \dots\dots\dots$

Visualized Solution

Given Information

The Chain Rule Strategy

Applying Chain Rule

Substituting $f^{'} (x)$

Updating $\frac{d y}{d x}$

The Quotient Rule Strategy

Quotient Rule Setup

Calculating Atomic Derivatives

Substituting Derivatives Back

Expanding the Numerator

Simplifying the Numerator

Final Result

The Sigma Insight: Techniques of Differentiation

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=f(x2+12x−1​) and f′(x)=sinx2, then dxdy​=………