Answer
$$\frac{d\langle p\rangle}{dt} = \langle -\frac{\partial V}{\partial x}\rangle $$
Work Step by Step
$\text{We'll be using the following equations in our solution:}$
$$\frac{\partial \psi}{\partial t} = \frac{\iota \hbar}{2m}\frac{\partial^2\psi}{\partial x^2} - \frac{\iota}{\hbar}V \psi \qquad...(1)$$
$$\frac{\partial \psi^*}{\partial t} = -\frac{\iota \hbar}{2m}\frac{\partial^2\psi^*}{\partial x^2} + \frac{\iota}{\hbar}V \psi^* \qquad...(2)$$
$\text{I will be not putting the limits of integration to keep the
}$ $\text{solution clean.}$
$$\langle p\rangle = \int \psi^*\big(\frac{\hbar}{\iota}\frac{\partial}{\partial x}\big)\psi dx $$
$$\langle p\rangle = -\iota \hbar \int \psi^*\frac{\partial \psi }{\partial x}dx
$$
$$\frac{d\langle p\rangle}{dt} =\frac{d}{dt}\big(-\iota \hbar \int \psi^*\frac{\partial \psi }{\partial x}dx\big)$$
$$\frac{d\langle p\rangle}{dt} =-\iota \hbar \int \frac{\partial}{\partial t}\big(\psi^*\frac{\partial \psi }{\partial x}\big)dx$$
$\text{Consider,}$
$$\frac{\partial}{\partial t}\big(\psi^*\frac{\partial \psi }{\partial x}\big) = \frac{\partial \psi^*}{\partial t} \frac{\partial \psi}{\partial x} + \psi^* \frac{\partial}{\partial t}\frac{\partial \psi}{\partial x}$$
$\text{But,}$
$$\frac{\partial}{\partial t}\frac{\partial \psi}{\partial x} = \frac{\partial}{\partial x}\frac{\partial \psi}{\partial t}$$
$\text{Hence,}$
$$\frac{\partial}{\partial t}\big(\psi^*\frac{\partial \psi }{\partial x}\big) = \frac{\partial \psi^*}{\partial t} \frac{\partial \psi}{\partial x} + \psi^* \frac{\partial}{\partial x}\frac{\partial \psi}{\partial t}$$
$\text{Using equation }1 \text{ and } 2$,
$$\frac{\partial}{\partial t}\big(\psi^*\frac{\partial \psi }{\partial x}\big) = \big(-\frac{\iota \hbar}{2m}\frac{\partial^2\psi^*}{\partial x^2} + \frac{\iota}{\hbar}V \psi^* \big)\frac{\partial \psi}{\partial x} + \psi^* \frac{\partial}{\partial x}\big(\frac{\iota \hbar}{2m}\frac{\partial^2\psi}{\partial x^2} - \frac{\iota}{\hbar}V \psi\big)$$
$$\frac{\partial}{\partial t}\big(\psi^*\frac{\partial \psi }{\partial x}\big) = \big(-\frac{\iota \hbar}{2m}\frac{\partial^2\psi^*}{\partial x^2}\frac{\partial \psi}{\partial x} + \frac{\iota}{\hbar}V \psi^*\frac{\partial \psi}{\partial x} \big) + \big(\frac{\iota \hbar}{2m}\psi^*\frac{\partial^3\psi}{\partial x^3} - \frac{\iota}{\hbar} \psi^* \frac{\partial}{\partial x}(V \psi)\big)$$
$$\frac{\partial}{\partial t}\big(\psi^*\frac{\partial \psi }{\partial x}\big) = \frac{\iota \hbar}{2m}\big(\psi^*\frac{\partial^3\psi}{\partial x^3} - \frac{\partial^2\psi^*}{\partial x^2}\frac{\partial \psi}{\partial x}\big) + \frac{\iota}{\hbar}V \psi^*\frac{\partial \psi}{\partial x} - \frac{\iota}{\hbar} \psi^* \frac{\partial}{\partial x}(V \psi)$$
$\text{Upon Integrating the first term vanishes, Hence:}$
$$\frac{\partial}{\partial t}\big(\psi^*\frac{\partial \psi }{\partial x}\big) = \frac{\iota}{\hbar}V \psi^*\frac{\partial \psi}{\partial x} - \frac{\iota}{\hbar} \psi^* \frac{\partial}{\partial x}(V \psi)$$
$$\frac{\partial}{\partial t}\big(\psi^*\frac{\partial \psi }{\partial x}\big) = -\frac{\iota}{\hbar}\psi^*\psi \frac{\partial V}{\partial x}$$
$\text{The initial equation finally becomes:}$
$$\frac{d\langle p\rangle}{dt} =-\iota \hbar \int \big(-\frac{\iota}{\hbar}\psi^*\psi \frac{\partial V}{\partial x}\big)dx$$
$$\frac{d\langle p\rangle}{dt} = \int \big(\psi^* \frac{-\partial V}{\partial x} \psi \big)dx$$
$$\frac{d\langle p\rangle}{dt} = \langle -\frac{\partial V}{\partial x}\rangle $$
$\text{Hence Proved}$
