Question

If \(y=e^{3\log(2x+1)}\), then \(\frac{dy}{dx}=\)

• A. \(6e^{3\log(2x+1)}\)
• B. \(6\frac{e^{3\log(2x+1)}}{2x+1}\)
• C. \(\frac{e^{3\log(2x+1)}}{2x+1}\)
• D. \(\frac{e^{3\log(2x+1)}}{3(2x+1)}\)
• E. \((2x+1)e^{3\log(2x+1)}\)

Answer 1

The Correct Option is B

We are given the function: \[ y = e^{3\log(2x+1)} \]
Step 1: Simplify the given expression Using the property of logarithms \( e^{\log(a)} = a \), we can simplify the expression for \( y \): \[ y = (2x + 1)^3 \]
Step 2: Differentiate \( y \) with respect to \( x \) Now, we differentiate \( y = (2x + 1)^3 \) using the chain rule: \[ \frac{dy}{dx} = 3(2x + 1)^2 \cdot \frac{d}{dx}(2x + 1) \] Since \( \frac{d}{dx}(2x + 1) = 2 \), we get: \[ \frac{dy}{dx} = 3 \cdot 2 \cdot (2x + 1)^2 = 6(2x + 1)^2 \]Step 3: Simplify the final expression Thus, the derivative is: \[ \frac{dy}{dx} = \frac{6e^{3\log(2x+1)}}{2x+1} \]

The correct option is (B) : \(6\frac{e^{3\log(2x+1)}}{2x+1}\)

Answer 2

• We are given: \[y=e^{3\log(2x+1)}\]
• Using the property \(a \log b = \log b^a\), we can rewrite the equation as: \[y=e^{\log(2x+1)^3}\]
• Since \(e^{\log a} = a\), we have: \[y = (2x+1)^3\]
• Now, differentiate \(y\) with respect to \(x\): \[\frac{dy}{dx} = \frac{d}{dx} (2x+1)^3\]
• Apply the chain rule: \[\frac{dy}{dx} = 3(2x+1)^2 \cdot \frac{d}{dx} (2x+1)\]
• The derivative of \((2x+1)\) is 2: \[\frac{dy}{dx} = 3(2x+1)^2 \cdot 2\]
• Simplify: \[\frac{dy}{dx} = 6(2x+1)^2\]
• Recall that \(y = (2x+1)^3\), so \((2x+1)^2 = \frac{y}{2x+1}\)^(2/3). Rewrite y as y = e^{3log(2x+1)}. So (2x+1)^2 = (e^{3log(2x+1)})^(2/3) = e^{2log(2x+1)}
• Rewrite dy/dx: \[\frac{dy}{dx} = 6e^{2log(2x+1)}\]
• The original answer choices do not contain 6e^{2log(2x+1)}. Going back to the initial result, \[\frac{dy}{dx} = 6(2x+1)^2 \]
• Rewrite the derivative in terms of \(y=e^{3\log(2x+1)}\): Since \(y = (2x+1)^3\) then \((2x+1)^2 = y^{\frac{2}{3}} = (e^{3\log(2x+1)})^{\frac{2}{3}} = e^{2\log(2x+1)}\). So, \[\frac{dy}{dx} = 6e^{2\log(2x+1)} \] This is not a correct choice. We must choose amongst the existing choices. Let's rework it from step 1. \[ \frac{dy}{dx} = e^{3\log(2x+1)} * 3 * \frac{2}{2x+1} = \frac{6e^{3\log(2x+1)}}{2x+1} \]
• Therefore, \(\frac{dy}{dx} = \frac{6e^{3\log(2x+1)}}{2x+1}\).