Convert $\partial t$ into $\partial x$ in a heat equation.

Question

We shall consider functions $h=h(t, x):[0, \infty) \times \mathbb{R} \rightarrow(0, \infty)$ which are $2 \pi$-periodic with respect to $x$, belong to $C^{\infty}([0, \infty) \times \mathbb{R})$ and satisfy the $1 \mathrm{D}$ heat equation $$ h_t-h_{xx}=0, \quad \text { in }(0, \infty) \times \mathbb{R} . $$ and the Boltzmann’s entropy is introduced: $$ H(t)=\int_0^{2\pi} h(t,x)\log(h(t,x))\mathrm{d}x.$$ The first task is to show that $\dfrac{\mathrm{d}H}{\mathrm{d}t}\leqslant 0$. My solution is as follow: \begin{align*} \dfrac{\mathrm{d}H}{\mathrm{d}t}&= \dfrac{\mathrm{d}}{\mathrm{d}t}\int_0^{2\pi} h(t,x)\log(h(t,x))\mathrm{d}x =\int_0^{2\pi} \dfrac{\mathrm{d}}{\mathrm{d}t}[h(t,x)\log(h(t,x))]\mathrm{d}x\\ &=\int_0^{2\pi}[h_t(\log(h)+1)]\mathrm{d}x=\int_0^{2\pi}[h_{xx}(\log(h)+1)]\mathrm{d}x\\ &=\int_0^{2\pi}(\log(h)+1)\mathrm{d}h_x =(\log h+1)h_x\bigg|_0^{2\pi}-\int_0^{2\pi}\frac{h^2_x}{h}\mathrm{d}x\\ &=-\int_0^{2\pi}\frac{h^2_x}{h}\mathrm{d}x\leqslant 0. \end{align*} But the formal answer is like: \begin{aligned} \frac{d}{d t} H(t) & =\int_0^{2 \pi} \partial_t[h(t, x) \log (h(t, x))] d x \\ & =\int_0^{2 \pi} \partial_x[(\log (h(t, x))+1) \partial h(t, x)]-\frac{\left|\partial_x h(t, x)\right|^2}{h(t, x)} d x \end{aligned} Can anyone see how to convert $\partial t$ into $\partial x$?