Question

If \[ (3y)^{2x}=5\left(2^{3x}\right), \] then \[ \left(\frac{dy}{dx}\right)_{x=1} = \]

• A. \(-\dfrac{\sqrt{10}\log 5}{3}\)
• B. \(-\dfrac{\sqrt{10}}{3}\)
• C. \(\dfrac{\sqrt{10}\log 5}{3}\)
• D. \(\dfrac{\sqrt{10}}{3}\)

Hint:

For equations of the form \(f(x,y)^{g(x)}=h(x)\), always use logarithmic differentiation. It converts exponents into products, making differentiation much easier.

Answer 1

The Correct Option is A

Concept: Whenever variables occur both in the base and exponent, logarithmic differentiation is the most efficient technique. We first take logarithms on both sides, then differentiate implicitly.

Step 1: Take logarithm on both sides. Given \[ (3y)^{2x}=5\cdot2^{3x}. \] Taking natural logarithm, \[ \log\left((3y)^{2x}\right) = \log\left(5\cdot2^{3x}\right). \] Using logarithmic laws, \[ 2x\log(3y) = \log5+3x\log2. \]

Step 2: Differentiate implicitly. Differentiating both sides with respect to \(x\), \[ 2\log(3y) + 2x\left(\frac{1}{y}\frac{dy}{dx}\right) = 3\log2. \] Hence, \[ \frac{2x}{y}\frac{dy}{dx} = 3\log2 - 2\log(3y). \] Therefore, \[ \frac{dy}{dx} = \frac{y}{2x} \Bigl[ 3\log2 - 2\log(3y) \Bigr]. \]

Step 3: Find \(y\) when \(x=1\). Substitute \(x=1\) into the original equation: \[ (3y)^2 = 5(2^3) = 40. \] Thus, \[ 3y=\sqrt{40}=2\sqrt{10}. \] Hence, \[ y=\frac{2\sqrt{10}}{3}. \]

Step 4: Substitute into derivative formula. At \(x=1\), \[ \frac{dy}{dx} = \frac{y}{2} \Bigl[ 3\log2 - 2\log(2\sqrt{10}) \Bigr]. \] Now, \[ 2\log(2\sqrt{10}) = 2\log2+\log10. \] Therefore, \[ 3\log2 - 2\log(2\sqrt{10}) = 3\log2-2\log2-\log10. \] \[ = \log2-\log10. \] \[ = \log\left(\frac{2}{10}\right) = -\log5. \] Thus, \[ \left(\frac{dy}{dx}\right)_{x=1} = \frac{1}{2} \left(\frac{2\sqrt{10}}{3}\right) (-\log5). \] Hence, \[ \boxed{ \left(\frac{dy}{dx}\right)_{x=1} = -\frac{\sqrt{10}\log5}{3} }. \]