Calculus & Geometric Derivatives

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$
Provided this two-sided limit exists.

It represents the average rate of change or the slope of the secant line passing through points $P(x,f(x))$ and $Q(x+h,f(x+h))$ on the function curve.

Point $Q$ slides along the curve towards point $P$. The secant line pivots around $P$ and gradually aligns with the tangent line at $P$, whose slope is the derivative $f^{\prime }(x)$.

$$\frac{d}{dx}\left[x^n\right]=n{x}^{n-1}$$
For any real number $n$. For example, $\frac{d}{dx}\left[x^3\right]=3x^2$ and $\frac{d}{dx}\left[x^2\right]=2x$.

$$\frac{d}{dx}[f(x)g(x)]=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$$
The derivative of the first function times the second, plus the first function times the derivative of the second.

$$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{[g(x){]}^2}$$
Provided $g(x)\ne 0$.

$$\frac{d}{dx}[f(g(x))]=f^{\prime }(g(x))\cdot g^{\prime }(x)$$
In Leibniz notation with $y=f(u)$ and $u=g(x)$: $$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$$.

The tangent line to the curve is horizontal (slope = 0). This indicates a critical point where the function may achieve a local maximum, local minimum, or horizontal inflection point.

No. Continuity does not imply differentiability. A classic counterexample is $f(x)=|x|$, which is continuous at $x=0$ but has a sharp corner where left and right limits disagree (not differentiable).