$L^+$ - limit from above
-$\lim_{x \to a} f(x)$ - limit of a point
+$\lim_{x \to a} f(x)$ - limit of a point
- Limit exists if $L^-=L^+$
- If limit exists, point does not.
## First principles derivative
-$$\lim_{\delta x \rightarrow 0}{\delta y \over \delta x}={dy \over dx} = f^\prime(x)$$
+$$f^\prime(x) = \lim_{\delta x \rightarrow 0}{\delta y \over \delta x}={dy \over dx}$$
$$m_{\operatorname{tangent}}=\lim_{h \rightarrow 0}f^\prime(x)$$
$$m_{\operatorname{chord PQ}}=f^\prime(x)$$
first principles derivative:
-$${m_{\operatorname{tangent at P}} =\lim_{h \rigzhtarrow 0}}{{f(x+h)-f(x)}\over h}$$
+$${m_{\operatorname{tangent at P}} =\lim_{h \rightarrow 0}}{{f(x+h)-f(x)}\over h}$$
+## Gradient at a point
+Given point $P(a, b)$ and function $f(x)$, the gradient is $f^\prime(a)$
+## Derivatives of $x^n$
+
+For $f: \mathbb{R} \rightarrow \mathbb{R}$ where $f(x)=x^n, x \in \mathbb{N}$
+
+Derivative is $f^\prime(x) = nx^{n-1}$
+
+If $x=$ constant, derivative is $0$
+
+If $f(x)={1 \over x}=x^{-1}, \quad f^\prime(x)=-1x^{-2}={-1 \over x^2}$
+
+If $f(x)=^5\sqrt{x}=x^{1 \over 5}, \quad f^\prime(x)={1 \over 5}x^{-4/5}={1 \over 5 \times ^5\sqrt{x^4}}$
+
+If $f(x)=(x-b)^2, \quad f^\prime(x)=2(x-b)$
+
+$$f^\prime(x)=\lim_{h \rightarrow 0}{{f(x+h)-f(x)} \over h}$$
+$$=\lim_{h \rightarrow 0}
+
## Euler's number as a limit
$$\lim_{h \rightarrow 0} {{e^h-1} \over h}=1$$