Question

1. y = \sqrt{e^{4x}-4x}

r 2(e^{4x}-4x)^{-\frac{1}{2}}(e^{4x}-1)
e (e^{4x}-4x)^{-\frac{1}{2}}(e^{4x}-1)
l e^{4x}(e^{4x}-4x)^{-\frac{1}{2}}
s 2e^{4x}(e^{4x}-4x)^{-\frac{1}{2}}

Explanation:

Step1: Simplify the function

Given \(y = \sqrt{e^{4x}-4x}=(e^{4x}-4x)^{\frac{1}{2}}\).

Step2: Use the chain - rule

The chain - rule states that if \(y = f(g(x))\), then \(y^\prime=f^\prime(g(x))\cdot g^\prime(x)\). Let \(u = e^{4x}-4x\), so \(y = u^{\frac{1}{2}}\). First, find the derivative of \(y\) with respect to \(u\): \(\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}\). Then find the derivative of \(u\) with respect to \(x\): \(u^\prime=\frac{d}{dx}(e^{4x}-4x)=4e^{4x}-4 = 4(e^{4x}-1)\).

Step3: Apply the chain - rule formula

By the chain - rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\). Substitute \(\frac{dy}{du}=\frac{1}{2}(e^{4x}-4x)^{-\frac{1}{2}}\) and \(\frac{du}{dx}=4(e^{4x}-1)\) into the formula: \(\frac{dy}{dx}=\frac{1}{2}(e^{4x}-4x)^{-\frac{1}{2}}\cdot4(e^{4x}-1)=2(e^{4x}-4x)^{-\frac{1}{2}}(e^{4x}-1)\).

Answer:

\(2(e^{4x}-4x)^{-\frac{1}{2}}(e^{4x}-1)\) (corresponds to option R)