1. The problem is to find the adjoint of the operator $\hat{A} = x \frac{d}{dx}$ using integration by parts, starting from the inner product: $$\langle \phi | \hat{A} \psi \rangle = \int \phi^* x \frac{d\psi}{dx} \, dx$$ 2. Recall the integration by parts formula: $$\int u \, dv = uv - \int v \, du$$ Here, let: $$u = \phi^* x, \quad dv = \frac{d\psi}{dx} dx$$ Then: $$du = \left( \frac{d\phi^*}{dx} x + \phi^* \right) dx, \quad v = \psi$$ 3. Applying integration by parts: $$\int \phi^* x \frac{d\psi}{dx} dx = \left[ \phi^* x \psi \right] - \int \psi \left( \frac{d\phi^*}{dx} x + \phi^* \right) dx$$ 4. Assuming boundary terms vanish (wavefunctions vanish at boundaries), the first term is zero: $$\left[ \phi^* x \psi \right] = 0$$ 5. So we have: $$\langle \phi | \hat{A} \psi \rangle = - \int \psi \left( x \frac{d\phi^*}{dx} + \phi^* \right) dx$$ 6. Taking complex conjugate inside the integral and rearranging: $$\langle \phi | \hat{A} \psi \rangle = \int \psi \left( -x \frac{d}{dx} - 1 \right) \phi^* dx = \langle \hat{A}^\dagger \phi | \psi \rangle$$ 7. Therefore, the adjoint operator is: $$\hat{A}^\dagger = -x \frac{d}{dx} - 1$$ Final answer: $$\boxed{\hat{A}^\dagger = -x \frac{d}{dx} - 1}$$