A simple application of the Fundamental Theorem of Calculus (FTC) as a reminder: $$\frac{\partial}{\partial x}\int_x^\infty f(t)dt=-f(x)$$
What if the integrand $f$ also depends on $x$?
So $$\frac{\partial}{\partial x}\int_x^\infty f(t,x)dt$$ then how do we apply the FTC?
Could it be that it just equals to $-f(x,x)$?
—- Solution according to @P. Lawrence's answer—-
Applying the Leibniz's Rule as suggested and correcting derivative notation gives
$$\frac{d}{d x}\int_x^\infty f(t,x)dt=\lim_{b\rightarrow\infty}\int_x^\infty \frac{\partial}{\partial x}f(t,x)dt-f(x,x)$$
Best Answer
$$\int_x^{\infty}f(t,x)dt$$ is a function just of $x$, not of $t$, which is only a dummy variable, living only inside the integral.
You should write $\frac{d}{dx}$, not $\frac{\partial }{\partial x}$ and you need Leibniz's Rule $$\frac{d}{dx}\int_{u(x)}^{v(x)}f(t,x)dt=\int_{u(x)}^{v(x)}\frac{\partial}{\partial x}f(t,x)dt+f(v(x),x)v'(x)-f(u(x),x)u'(x)$$ Now put $$u(x)=x\\v(x)=b$$ so $$u'(x)=1\\v'(x)=0$$ and take the limit as $b \rightarrow \infty $, provided the limit exists.