How do I find $\frac{\mathrm{d}}{\mathrm{d}x} \int_{x}^{\infty} f(t, x) \mathrm{d} t$?

I am trying to compute the following derivative:

$$\frac{\mathrm{d}\Big(\int_{x}^{\infty} f(t, x) \mathrm{d} t\Big)}{\mathrm{d}x} \text{.}\tag{1}$$

It is straightforward to compute the following derivative using the Fundamental Theorem of Calculus:

$$\frac{\mathrm{d}\Big(\int_{x}^{\infty} f(t) \mathrm{d} t\Big)}{\mathrm{d}x} = - f(x)\text{,}\tag{2}$$

but I am not sure how to change this if $x$ appears in the integrand.

I attempted to use wolframalpha.com, and received the following output:

$$\frac{\mathrm{d}\Big(\int_{t=x}^{\infty} f(t, x) \mathrm{d} t\Big)}{\mathrm{d}x} = \int_{x}^{\infty} f^{(0, 1)}(t, x) \mathrm{d} t - f(x, x)\text{,}\tag{3}$$

but I do not know the meaning of $f^{(0, 1)}(t, x)$ or where this solution comes from.

Edit: You can assume $f$ satisfies reasonable smoothness properties.

$f^{(0,1)}$ in WolphramAlpha means the partial derivative over the second argument, that is $$ f^{(0,1)}(t,x) = \frac{\partial f}{\partial x}(t,x)$$ The result that was given to you comes from the Leibniz integral rule.