For x 1, if (2x)^(2y) = 4e^(2x-2y), then (1 + log_e 2x)² dy/dx is equal to : (A) log_e 2x (B) (x l

Q: Wait, why can't I just use implicit differentiation on the original equation $(2x)^{2y} = 4e^{2x-2y}$?

You absolutely can! However, implicit differentiation here would involve the product rule and chain rule on both sides, leading to a very messy expression. By taking the natural log first, we transform the exponential equation into a simple algebraic one, making the differentiation process much cleaner and less prone to calculation errors.

Q: Why did $\ln 4$ become $2 \ln 2$ in the steps?

That is a great observation! We use the logarithmic property $\ln(a^b) = b \ln a$. Since $4 = 2^2$, $\ln 4$ becomes $\ln(2^2)$, which is $2 \ln 2$. This simplification is crucial because it allows us to factor out the constant $2$ from the entire equation, making the final expression for $y$ much easier to handle.

Q: Is there a trick to the final simplification where the $1$s cancel out?

It is not a trick, but rather the natural result of the algebra! When you apply the quotient rule, you get a term involving $(1 + \ln 2x)$ and a term involving $(x + \ln 2) \cdot \frac{1}{x}$. When you expand the latter, you get $1 + \frac{\ln 2}{x}$. The $1$ from the first part and the $1$ from the expansion cancel out perfectly, leaving you with the elegant final result. Always trust the algebra!

Question

For $x > 1$, if $(2x)^{2y} = 4e^{2x-2y}$, then $(1 + \log_e 2x)^2 \frac{dy}{dx}$ is equal to :

Options:

(A) $\log_e 2x$

(B) $\frac{x \log_e 2x + \log_e 2}{x}$

(C) $x \log_e 2x$

(D) $\frac{x \log_e 2x - \log_e 2}{x}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex equation:

$$(2x)^{2y} = 4e^{2x-2y}$$

At first glance, it looks intimidating. The variable $y$ is trapped in the exponent, and the base $2x$ is a function of $x$.

In the world of calculus, we use the natural logarithm as our master key. By taking the natural log of both sides and invoking the power rule $\ln(a^b) = b \ln a$, we bring the trapped $y$ down to the ground level.

The equation transforms into:

$$2y \ln(2x) = \ln(4) + (2x - 2y) \ln(e)$$

Since $\ln(e) = 1$, the equation simplifies to:

$$2y \ln(2x) = \ln(4) + 2x - 2y$$

### The Algebraic Dance: Isolating the Variable

Now that we have brought $y$ down, our next mission is to isolate it. We group all terms containing $y$ on one side:

$$2y \ln(2x) + 2y = 2x + \ln(4)$$

Factoring out $2y$ gives us:

$$2y(\ln(2x) + 1) = 2x + 2 \ln(2)$$

Note that $\ln(4)$ is equivalent to $2 \ln(2)$. Dividing the entire equation by $2$, we obtain:

$$y(1 + \ln(2x)) = x + \ln(2)$$

Finally, we isolate $y$ to get:

$$y = \frac{x + \ln(2)}{1 + \ln(2x)}$$

### The Calculus Crucible: Applying the Quotient Rule

With $y$ isolated, we differentiate using the quotient rule:

$$\frac{dy}{dx} = \frac{v u' - u v'}{v^2}$$

Here, $u = x + \ln(2)$ and $v = 1 + \ln(2x)$. The derivative of the numerator $u$ is $1$, as $\ln(2)$ is a constant.

The derivative of the denominator $v$ is:

$$\frac{d}{dx}(1 + \ln(2x)) = \frac{1}{2x} \cdot 2 = \frac{1}{x}$$

Assembling these into the quotient rule, we get:

$$\frac{dy}{dx} = \frac{(1 + \ln(2x))(1) - (x + \ln(2))(\frac{1}{x})}{(1 + \ln(2x))^2}$$

### The Final Reveal: Elegance in Cancellation

The problem asks for the value of $(1 + \ln(2x))^2 \frac{dy}{dx}$. By multiplying our derivative by the denominator squared, we cancel the denominator:

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

Expanding this expression:

$$1 + \ln(2x) - \left( \frac{x}{x} + \frac{\ln(2)}{x} \right)$$

This simplifies to:

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

The $1$ and $-1$ cancel out, leaving us with $\ln(2x) - \frac{\ln(2)}{x}$. Finding a common denominator, we arrive at the final result:

$$\mathbf{\frac{x \ln(2x) - \ln(2)}{x}}$$

Animated Solution for Mathematics - Differentiation: For $x > 1$ , if $(2 x)^{2 y} = 4 e^{2 x - 2 y}$ , then $(1 + lo g_{e} 2 x)^{2} \frac{d y}{d x}$ is equal to :

Select Answer:

Visualized Solution

Analyze the Given Equation

Apply Natural Logarithm

Simplify Using Log Properties

Group Terms with $y$

Factor Out $y$

Isolate $y$ as a Function of $x$

Apply the Quotient Rule

Differentiate the Components

Multiply by the Denominator Squared

Final Algebraic Simplification

Conclusion and Final Answer

The Sigma Insight: Techniques of Differentiation

Analyzing the Setup

The Algebraic Dance

The Calculus Crucible

The Final Reveal

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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; } For x>1, if (2x)2y=4e2x−2y, then (1+loge​2x)2dxdy​ is equal to :

Select Answer:

Visualized Solution

Analyze the Given Equation

Apply Natural Logarithm

Simplify Using Log Properties

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

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

Isolate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y as a Function of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x