Question

\(y=x \ln x+\sin (\pi \ln x)\) \(\frac{d y}{d x}=1+\ln x+\cos (\pi \ln x) \frac{\pi}{x}\) at \(x=e\) \(\frac{d y}{d x}=2-\frac{\pi}{e}=\frac{2 e-\pi}{e}\) at \(x=e^{2}\) \(\frac{d y}{d x}=3+\frac{\pi}{e^{2}}=\frac{3 e^{2}+\pi}{e^{2}}\)

Short answer

Answer: At \(x=e\), the derivative of the function is \(\frac{dy}{dx}= \frac{2e-\pi}{e}\), and at \(x=e^2\), the derivative is \(\frac{dy}{dx}= \frac{3e^2+\pi}{e^2}\).

Step-by-step solution

Find the derivative of y with respect to x

First, we need to differentiate the given function with respect to \(x\). Recall that you can differentiate each term separately using the Chain Rule, Power Rule, and Product Rule. For \(x \ln x\): 1. Differentiate \(x\): \(\frac{dx}{dx} = 1\) 2. Differentiate \(\ln x\): \(\frac{d(\ln x)}{dx} = \frac{1}{x}\) 3. Apply the Product Rule: \(\frac{d(x \ln x)}{dx} = x\left(\frac{1}{x}\right) + \ln x (1) = 1+\ln x\) For \(\sin(\pi \ln x)\): 1. Differentiate \(\sin(\pi \ln x)\): \(\frac{d(\sin (\pi \ln x))}{dx} = \cos(\pi \ln x) \cdot \frac{d(\pi \ln x)}{dx}\) 2. Differentiate \(\pi \ln x\): \(\frac{d(\pi \ln x)}{dx} = \pi \frac{d(\ln x)}{dx} = \pi \cdot \frac{1}{x}\) 3. Combine the terms: \(\frac{d(\sin (\pi \ln x))}{dx} = \cos(\pi \ln x) \cdot \frac{\pi}{x}\) Now, sum the derivatives of each term: \(\frac{dy}{dx} = (1 + \ln x) + \cos(\pi \ln x)\frac{\pi}{x}\)

Evaluate the derivative at x=e and x=e^2

Now, we will evaluate the derivative at \(x=e\) and \(x=e^2\). At \(x=e\): \(\frac{dy}{dx} = \left(1+\ln e\right) + \cos(\pi \ln e) \frac{\pi}{e} = 1+1+\cos(\pi) \frac{\pi}{e} = 2-\frac{\pi}{e} = \frac{2e-\pi}{e}\) At \(x=e^2\): \(\frac{dy}{dx} = \left(1+\ln e^2\right) + \cos(\pi \ln e^2) \frac{\pi}{e^2} = 1+2+\cos(2\pi) \frac{\pi}{e^2} = 3+\frac{\pi}{e^2} = \frac{3e^2+\pi}{e^2}\) Therefore, at \(x=e\), the derivative is \(\frac{dy}{dx}= \frac{2e-\pi}{e}\) and at \(x=e^2\), the derivative is \(\frac{dy}{dx}= \frac{3e^2+\pi}{e^2}\).