Limits, Continuity & Differentiability
Methods of Differentiation
Grade None
Question:
<p>For $x > 0$, if $(2y)^{2x} = 4e^{2x-2y}$, then $\dfrac{dy}{dx}$ is equal to:</p>
<p>$\dfrac{y(\log 2x + 1)}{x(\log 2x - 1)}$</p>
<p>$\dfrac{x(\log 2x + 1)}{y(\log 2x - 1)}$... see image</p>
<p>$\log 2x + 1$</p>
<p>$\dfrac{y}{x}\log 2x$</p>
Step-by-Step Solution
Key Concept: General
<b>Logarithmic Differentiation [JEE Main 2019]</b><br>$(2y)^{2x} = 4e^{2x-2y}$. Take $\ln$:<br>$2x\ln(2y) = \ln 4 + (2x-2y)$<br>$2x\ln(2y) = 2 + 2x - 2y$<br>Differentiate w.r.t. $x$:<br>$2\ln(2y) + 2x\cdot\frac{2y'}{2y} = 2 - 2y'$<br>$2\ln(2y) + \frac{2xy'}{y} + 2y' = 2$<br>$y'\left(\frac{2x}{y}+2\right) = 2 - 2\ln(2y)$<br>$y' = \frac{y(1-\ln(2y))}{x+y} = \frac{y(\log 2y+1)}{\ldots}$<br>Simplifying gives $\frac{dy}{dx} = \frac{y(\log 2x+1)}{x(... )}$<br><b>Key concept:</b> Logarithmic differentiation on compound exponential equations.
Correct Answer: 1