I saw a result a while ago which stated if $f(Y) = \int^{Y} g(t)dt$, then $f'(Y) = g(Y)$, where we differentiate with respect to Y. I'm unable to find any sources on this result (in particular, what it is called). I believe it is used in this question
What is the name of this result?
This result is the first part of the fundamental theorem of calculus. Let $f$ be a function that is continuous on the interval $[a,b]$. Then, let $F:[a,b] \mapsto \mathbb{R}$ be the function defined by $$ F(x)=\int_{a}^{x}f(t) \, dt $$ The fundamental theorem of calculus states that $F$ is continuous on $[a,b]$, and $$ F'(x)=f(x) $$ for all $x$ in $(a,b)$. This result can be summarised as $$ \frac{d}{dx}\int_{a}^{x}f(t) \, dt = f(x) \, . $$ Integrating a function, followed by differentiating it, gets you back where you started. Note that integrating refers to finding the area under the curve. Really, that's what an integral is. The fact that functions of the form $$ \int f(x) \, dx $$ are often called 'indefinite integrals' is a result of the fundamental theorem of calculus, since $$ \int_{a}^{x}f(t) \, dt = \int f(x) \, dx \, . $$
