Core Ideas in Plain Terms

The derivative $f'(x)$ is the slope of the graph right at a point — how fast the output is changing at that exact instant. If $f$ is your position over time, $f'$ is your speed. The power rule $(x^n)' = nx^{n-1}$ is the fast way to compute it for powers.

An integral $\int_a^b f(x)\,dx$ adds up infinitely many thin slices to get a total — most often the area under the curve between $a$ and $b$. If $f$ is your speed over time, the integral is the total distance you travelled.

The Fundamental Theorem of Calculus says that integrating and differentiating undo each other. To find $\int_a^b f$, find a function $F$ whose derivative is $f$, then just compute $F(b) - F(a)$ — no adding up tiny slices by hand.

If $y$ changes $3$ times as fast as $u$, and $u$ changes $2$ times as fast as $x$, then $y$ changes $6$ times as fast as $x$. That is the whole chain rule: $\dfrac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$ — differentiate the outside, keep the inside, then multiply by the inside's rate.

Slice the region under a curve into thin vertical strips. Each strip is nearly a rectangle: height $f(x)$, tiny width. The definite integral $\int_a^b f(x)\,dx$ is what those rectangle areas add up to as the strips get thinner — which is also why integrating a rate (like speed) recovers a total (like distance). That derivative–integral link is the Fundamental Theorem of Calculus: $\int_a^b F'(x)\,dx = F(b) - F(a)$.

Before touching symbols, say what the question IS: a rate or slope → differentiate; an accumulated total or area → integrate. And always check an integral by differentiating your antiderivative — it should give back the integrand.

1. Plug in. If nothing breaks, that's the limit. 2. Factor and cancel when plug-in gives $\tfrac00$ — both parts share the root, so divide it out. 3. Multiply by the conjugate when a square root blocks the factoring. 4. L'Hôpital's rule is a way: on a genuine $\tfrac00$ or $\tfrac\infty\infty$ form, take the derivative of the top and the bottom separately and try again. 5. Squeeze an oscillating expression between two bounds that agree.

The rule needs an actual $\tfrac00$ or $\tfrac{\infty}{\infty}$. Using it on $\lim_{x\to0}\tfrac{x+1}{x+2} = \tfrac12$ (not an indeterminate form) gives the WRONG answer $1$. Check the form first, every time.