If y = y(x) is an implicit function of x such that logₑ(x + y) = 4xy, then d²y/dx² at x = 0 is equal

Q: Wait, why do we convert to exponential form? Can't we just differentiate $\log_e(x + y) = 4xy$ directly?

You absolutely could, but you would be walking into a trap! Differentiating $\log_e(x + y)$ directly forces you to use the chain rule, resulting in $\frac{1}{x+y} \cdot (1 + y')$. This introduces a fraction that makes the subsequent algebra for the second derivative significantly more complex and prone to calculation errors. Converting to $x + y = e^{4xy}$ is a strategic move that keeps the equation in a polynomial-like structure, making the product and chain rules much cleaner to apply.

Q: Why do we need to find $y$ and $y'$ at $x = 0$ before finding $y''$?

Think of this as 'clearing the path' before the final climb. If you try to find the general expression for $y''$ without substituting the numerical values for $y$ and $y'$ at $x=0$ along the way, you will end up with an algebraic nightmare. By solving for $y$ and $y'$ at $x=0$ first, you reduce the complexity of the second derivative equation, allowing you to focus on the calculus rather than getting lost in a sea of variables.

Q: I keep getting confused with the product rule inside the chain rule. How do I handle that?

It is completely normal to feel that way! When you differentiate $e^{4xy}$, treat the exponent $4xy$ as a separate function, say $u = 4xy$. The derivative is $e^u \cdot \frac{du}{dx}$. When you calculate $\frac{du}{dx}$, that is where the product rule lives: $\frac{d}{dx}(4xy) = 4(x \cdot y' + y \cdot 1)$. Just take it one layer at a time. Don't try to do the whole thing in your head; write out the derivative of the exponent separately, then plug it back into the chain rule structure.

Question

If $y = y(x)$ is an implicit function of $x$ such that $\log_e(x + y) = 4xy$, then $\frac{d^2y}{dx^2}$ at $x = 0$ is equal to .

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### Analyzing the Setup

Imagine you are standing at the edge of a complex mathematical landscape. You are presented with the equation $\log_e(x + y) = 4xy$, and your mission is to find the second derivative, $\frac{d^2y}{dx^2}$, at the point $x = 0$.

At first glance, this looks like a daunting task. The logarithm is intertwined with an implicit function, and the prospect of differentiating this twice seems like a recipe for an algebraic headache.

But fear not! In this masterclass, we will peel back the layers of this problem, transforming a terrifying expression into a series of elegant, manageable steps.

### Phase 1: The Liberation

The first step is often the most important. We are given $\log_e(x + y) = 4xy$.

If we differentiate this as it stands, we are forced to deal with the derivative of a logarithm, which introduces a fraction: $\frac{1}{x+y} \cdot (1 + y')$. This is a trap.

Instead, let us use the fundamental property of logarithms: $\log_e(A) = B \Rightarrow A = e^B$. By rewriting our equation as $x + y = e^{4xy}$, we have liberated ourselves from the fraction.

This exponential form is much friendlier to the chain rule. Before we proceed, we need to know where we are on the graph.

At $x = 0$, our equation becomes $0 + y = e^{4(0)y}$, which simplifies to $y = e^0$, or $y = 1$. So, we are working at the point $(0, 1)$. Keep this coordinate in your pocket; it will be our anchor.

### Phase 2: The First Derivative

Now, we differentiate $x + y = e^{4xy}$ with respect to $x$. On the left, the derivative is simple: $1 + y'$.

On the right, we apply the chain rule. The derivative of $e^{4xy}$ is $e^{4xy}$ multiplied by the derivative of the exponent $4xy$. Using the product rule, the derivative of $4xy$ is $4(y + xy')$.

Thus, our first derivative equation is:
$$1 + y' = e^{4xy}(4y + 4xy')$$

Let us find the value of $y'$ at $x = 0$. Substituting $x = 0$ and $y = 1$, we get:
$$1 + y' = e^0(4(1) + 4(0)y')$$
$$1 + y' = 4$$

Therefore, $y' = 3$ at $x = 0$. We are making excellent progress.

### Phase 3: The Second Derivative

Now for the main event: the second derivative. We differentiate $1 + y' = e^{4xy}(4y + 4xy')$ again.

The left side becomes $y''$. The right side is a product of two functions: $u = e^{4xy}$ and $v = (4y + 4xy')$. Applying the product rule, we get $u'v + uv'$.

The first part, $u'v$, is:
$$e^{4xy}(4y + 4xy') \cdot (4y + 4xy') = e^{4xy}(4y + 4xy')^2$$

The second part, $uv'$, is $e^{4xy} \cdot \frac{d}{dx}(4y + 4xy')$. Differentiating the second term, we get $4y' + 4(y' + xy'') = 8y' + 4xy''$.

So, the full expression is:
$$y'' = e^{4xy}(4y + 4xy')^2 + e^{4xy}(8y' + 4xy'')$$

### Phase 4: The Collapse

This looks like a monster, but watch what happens when we substitute $x = 0, y = 1,$ and $y' = 3$. The $e^{4xy}$ terms become $e^0 = 1$.

The first term becomes:
$$1 \cdot (4(1) + 4(0)(3))^2 = 4^2 = 16$$

The second term becomes:
$$1 \cdot (8(3) + 4(0)y'') = 24$$

Adding these together, $y'' = 16 + 24 = 40$. The monster has collapsed into a simple, beautiful integer.

We have arrived at our destination: the second derivative at $x = 0$ is **40**. This journey shows that even the most intimidating calculus problems can be tamed with the right strategy and a calm, step-by-step approach.

Animated Solution for Mathematics - Differentiation: If $y = y (x)$ is an implicit function of $x$ such that $lo g_{e} (x + y) = 4 x y$ , then $\frac{d ^{2} y}{d x ^{2}}$ at $x = 0$ is equal to .

Enter Numerical Value:

Visualized Solution

Analyze the Implicit Function

Convert to Exponential Form

Find the Value of $y$ at $x = 0$

Differentiate Once Implicitly

Apply the Product Rule

Find $y^{'}$ at $x = 0$

Prepare for Second Derivative

Differentiate the RHS (Part 1)

Differentiate the RHS (Part 2)

The Complete Second Derivative Equation

Substitute Values at $x = 0$

Final Calculation

The Sigma Insight: Higher Order Derivatives

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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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=y(x) is an implicit function of x such that loge​(x+y)=4xy, then dx2d2y​ at x=0 is equal to .