If x^(log_e(log_e x)) - x² + y² = 4 (y 0), then dy/dx at x = e is equal to : (A) e/√(4+e²) (B) (1+

Q: Why can't I just differentiate and then plug in $x=e$? Why do I need $y$?

Implicit differentiation creates a relationship between $x$, $y$, and $\frac{dy}{dx}$. If you don't find $y$ first, your final answer will be stuck with an unknown variable $y$ that you cannot evaluate. Always find the full coordinate $(x, y)$ first!

Q: Why did we reject the negative root for $y$?

Great catch! The problem explicitly states $y > 0$. In mathematics, we must respect the constraints given in the problem statement. Always check the domain and conditions before finalizing your values.

Q: Is the derivative of $\log_e(\log_e x)$ just $\frac{1}{\log_e x}$?

Not quite! You missed the Chain Rule. The derivative of $\log_e(u)$ is $\frac{1}{u} \cdot u'$. Here, $u = \log_e x$, so $u' = \frac{1}{x}$. Thus, the derivative is $\frac{1}{\log_e x} \cdot \frac{1}{x}$. Don't forget that inner derivative!

Question

If $x^{\log_e(\log_e x)} - x^2 + y^2 = 4 (y > 0)$, then $dy/dx$ at $x = e$ is equal to :

Options:

(A) $\frac{e}{\sqrt{4+e^2}}$

(B) $\frac{1+2e}{2\sqrt{4+e^2}}$

(C) $\frac{2e-1}{2\sqrt{4+e^2}}$

(D) $\frac{1+2e}{4+e^2}$

Sigma Percentile · Accepted Answer

### The Calculus of Hidden Paths

Welcome, fellow traveler of the mathematical landscape. Today, we stand before a problem that looks like a tangled knot of logarithms and variables:

$$x \log_e(\log_e x) - x^2 + y^2 = 4$$

It is easy to feel overwhelmed when you see nested functions, but remember: calculus is not about brute force; it is about elegance and strategy. Let us embark on this journey to find the slope of this curve at $x=e$.

### Phase 1: The Detective Work

Before we even touch the derivative, we must act like detectives. We are asked for $\frac{dy}{dx}$ at $x=e$. If we jump straight into differentiation, we will end up with an expression containing both $x$ and $y$.

If we do not know the value of $y$ at $x=e$, we are essentially trying to solve a puzzle with a missing piece. So, let us find $y$ first.

Substitute $x=e$ into our original equation:

$$e \log_e(\log_e e) - e^2 + y^2 = 4$$

Now, take a deep breath. Look at the term $\log_e e$. We know that $\log_e e = 1$. This simplifies our expression to:

$$e \log_e(1) - e^2 + y^2 = 4$$

And what is $\log_e 1$? It is zero! The entire first term vanishes, leaving us with $-e^2 + y^2 = 4$. With the condition $y > 0$, we find:

$$y = \sqrt{4 + e^2}$$

We have our coordinate. The path is clear.

### Phase 2: The Calculus Engine

Now, we apply the operator $\frac{d}{dx}$ to the entire equation. This is implicit differentiation.

We treat $y$ as a function of $x$, which means every time we differentiate a term involving $y$, we must multiply by $\frac{dy}{dx}$ due to the Chain Rule. The equation becomes:

$$\frac{d}{dx} [x \log_e(\log_e x)] - \frac{d}{dx}[x^2] + \frac{d}{dx}[y^2] = \frac{d}{dx}[4]$$

### Phase 3: The Chain Rule Symphony

Let us focus on the most intimidating part: $\frac{d}{dx} [x \log_e(\log_e x)]$. This is a product of two functions, $u=x$ and $v=\log_e(\log_e x)$. The Product Rule tells us the derivative is $u'v + uv'$.

The derivative of $x$ is $1$. The derivative of $\log_e(\log_e x)$ requires the Chain Rule:

$$\frac{1}{\log_e x} \cdot \frac{d}{dx}(\log_e x) = \frac{1}{\log_e x} \cdot \frac{1}{x}$$

When we combine these, we get $1 \cdot \log_e(\log_e x) + x \cdot (\frac{1}{\log_e x} \cdot \frac{1}{x})$. Notice the beauty of the cancellation!

The $x$ in the numerator and denominator vanishes, leaving us with $\log_e(\log_e x) + \frac{1}{\log_e x}$. This is the elegance of calculus—complexity collapsing into simplicity.

### Phase 4: The Grand Finale

We assemble our pieces:

$$\log_e(\log_e x) + \frac{1}{\log_e x} - 2x + 2y \frac{dy}{dx} = 0$$

Now, we substitute $x=e$. Again, the logs simplify: $\log_e(\log_e e) = \log_e(1) = 0$, and $\frac{1}{\log_e e} = 1$. Our equation becomes:

$$0 + 1 - 2e + 2y \frac{dy}{dx} = 0$$

Solving for $\frac{dy}{dx}$, we get:

$$\frac{dy}{dx} = \frac{2e - 1}{2y}$$

Finally, we plug in our value for $y$ that we found in Phase 1:

$$\frac{dy}{dx} = \frac{2e - 1}{2\sqrt{4 + e^2}}$$

Look at that result. It is clean, precise, and correct. You have navigated the nested logs, handled the implicit differentiation, and respected the constraints. This is the essence of JEE Advanced mathematics—not just calculation, but the art of logical progression.

Animated Solution for Mathematics - Differentiation: If $x^{l o g_{e} (l o g_{e} x)} - x^{2} + y^{2} = 4 (y > 0)$ , then $d y / d x$ at $x = e$ is equal to :

Select Answer:

Visualized Solution

The Given Equation

The Objective

Finding $y$ at $x = e$

Simplifying the Logarithm

Calculating $y$

Implicit Differentiation Setup

Applying Product Rule

Differentiating the Nested Logarithm

Differentiating the Remaining Terms

The Complete Derivative Equation

Substituting $x = e$ into the Derivative

Simplifying the Derivative at $x = e$

Isolating the $\frac{d y}{d x}$ term

Final Substitution of $y$

The Sigma Insight: Techniques of Differentiation

The Calculus of Hidden Paths

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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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 xloge​(loge​x)−x2+y2=4(y>0), then dy/dx at x=e is equal to :

Select Answer:

Visualized Solution

The Given Equation

The Objective

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

Simplifying the Logarithm

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Implicit Differentiation Setup

Applying Product Rule

Differentiating the Nested Logarithm

Differentiating the Remaining Terms

The Complete Derivative Equation

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=e into the Derivative

Simplifying the Derivative at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=e

Isolating the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdy​ term

Final Substitution of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y