7. maximum mark: 4 consider the function defined by ( f(x)=e^{4x}sin(2x) ). find ( fleft(\frac{pi}{4}\right)…

Explanation:

Step1: Use the product rule

The product rule states that if (y = u\cdot v), then (y^\prime=u^\prime v + uv^\prime). Let (u = e^{4x}) and (v=\sin(2x)). First, find (u^\prime): Using the chain rule, if (u = e^{4x}), then (u^\prime=\frac{d}{dx}(4x)\cdot e^{4x}=4e^{4x}). Next, find (v^\prime): Using the chain rule, if (v=\sin(2x)), then (v^\prime=\frac{d}{dx}(2x)\cdot\cos(2x) = 2\cos(2x)). So, (f^\prime(x)=4e^{4x}\sin(2x)+2e^{4x}\cos(2x)).

Step2: Substitute (x = \frac{\pi}{4})

Substitute (x=\frac{\pi}{4}) into (f^\prime(x)): (\sin\left(2\times\frac{\pi}{4}\right)=\sin\left(\frac{\pi}{2}\right) = 1), (\cos\left(2\times\frac{\pi}{4}\right)=\cos\left(\frac{\pi}{2}\right)=0). (f^\prime\left(\frac{\pi}{4}\right)=4e^{4\times\frac{\pi}{4}}\times1+2e^{4\times\frac{\pi}{4}}\times0). Simplify the exponent: (4\times\frac{\pi}{4}=\pi). So, (f^\prime\left(\frac{\pi}{4}\right)=4e^{\pi}).

Answer:

(4e^{\pi})