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CALCULUS - SINGLE VARIABLE

Infinitesimal

x as infinite if it satisfies the conditions |x| > 1, |x| > 1 + 1, |x| > 1 + 1 + 1, ..., and infinitesimal if x ≠ 0 and a similar set of conditions holds for x and the reciprocals of the positive integers

Continuity and Limit

$$\lim _{x\to p}f(x)=L,$$

the function has a limit L at an input p, if f(x) gets closer and closer to L as x moves closer and closer to p

continuity of $y=f(x)$ by saying that an infinitesimal change in x necessarily produces an infinitesimal change in y

A function f is said to be continuous at c if it is both defined at c and its value at c equals the limit of f as x approaches c:

$$\lim _{x\to c}f(x)=f(c).$$ We have here assumed that c is a limit point of the domain of f.

$$(\forall \varepsilon >0)\,(\exists \delta >0)\,(\forall x\in \mathbb {R} )\,(0<|x-p|<\delta \implies |f(x)-L|<\varepsilon ).$$

$$\lim _{x\to \infty }f(x)=L,$$

Differential Calculus

$m={\frac {\text{rise}}{\text{run}}}={\frac {{\text{change in }}y}{{\text{change in }}x}}={\frac {\Delta y}{\Delta x}}.$

$$L=\lim _{h\to 0}{\frac {f(a+h)-f(a)}{h}}$$

For every positive real number ⁠$\varepsilon$⁠, there exists a positive real number $\delta$ such that, for every $h$ such that $|h|<\delta$ and $h\neq 0$ then $f(a+h)$ is defined, and $$\left|L-{\frac {f(a+h)-f(a)}{h}}\right|<\varepsilon ,$$

Leibniz Derivative Notation

The first derivative of $y=f(x)$ is denoted by ⁠$\textstyle {\frac {dy}{dx}}$⁠, read as "the derivative of $y$ with respect to ⁠$x$⁠".

${\textstyle {\frac {dy}{dx}}={\frac {d}{dx}}f(x).}$

${\textstyle {\frac {d^{n}y}{dx^{n}}}}$ for the $n$-th derivative of $y=f(x)$

${\textstyle {\frac {d^{2}y}{dx^{2}}}={\frac {d}{dx}}{\Bigl (}{\frac {d}{dx}}f(x){\Bigr )}.}$

Lagrange Derivative Notation

The first derivative is written as ⁠$f'(x)$⁠

$f^{(n)}$ for the ⁠$n$⁠th derivative of ⁠$f$⁠.

Newton Derivative Notation

If $y$ is a function of ⁠$t$⁠, then the first and second derivatives can be written as ${\dot {y}}$ and ⁠${\ddot {y}}$⁠

Differential Operator Derivative Notation

$Df(x)$ and higher derivatives are written with a superscript, so the $n$-th derivative is ⁠$D^{n}f(x)$⁠

Derivative Example

Let f(x) = x² be the squaring function.

${\begin{aligned}f'(3)&=\lim _{h\to 0}{(3+h)^{2}-3^{2} \over {h}}\\&=\lim _{h\to 0}{9+6h+h^{2}-9 \over {h}}\\&=\lim _{h\to 0}{6h+h^{2} \over {h}}\\&=\lim _{h\to 0}(6+h)\\&=6\end{aligned}}$

Leibniz Derivative Example

${\begin{aligned}y&=x^{2}\\{\frac {dy}{dx}}&=2x.\end{aligned}}$

${\frac {d}{dx}}(x^{2})=2x.$

Derivative Rules

Differentiation is linear

$h(x)=af(x)+bg(x)$ with respect to $x$ is: $h'(x)=af'(x)+bg'(x).$

• The constant factor rule $$(af)'=af'$$
• The sum rule $$(f+g)'=f'+g'$$
• The difference rule $$(f-g)'=f'-g'.$$

$${\frac {d(af+bg)}{dx}}=a{\frac {df}{dx}}+b{\frac {dg}{dx}}.$$

Constant Rule

$f(x)=c$, then ${\frac {df}{dx}}=0$

Power Rule

$f(x)=x^{r}$

$f'(x)=rx^{r-1}\,.$

Product Rule

For the functions $f$ and $g$, the derivative of the function $h(x)=f(x)g(x)$ with respect to $x$ is $$h'(x)=(fg)'(x)=f'(x)g(x)+f(x)g'(x).$$ In Leibniz's notation this is written $${\frac {d(fg)}{dx}}=g{\frac {df}{dx}}+f{\frac {dg}{dx}}.$$

Chain Rule

The derivative of the function $h(x)=f(g(x))$ is $$h'(x)=f'(g(x))\cdot g'(x).$$

In Leibniz's notation, this is written as: $${\frac {d}{dx}}h(x)=\left.{\frac {d}{dz}}f(z)\right|_{z=g(x)}\cdot {\frac {d}{dx}}g(x),$$ often abridged to $${\frac {dh(x)}{dx}}={\frac {df(g(x))}{dg(x)}}\cdot {\frac {dg(x)}{dx}}.$$

$$h'=(f\circ g)'=(f'\circ g)\cdot g'.$$

If z = f(y) and y = g(x)

$${\frac {dz}{dx}}={\frac {dz}{dy}}\cdot {\frac {dy}{dx}},$$ and $$\left.{\frac {dz}{dx}}\right|_{x}=\left.{\frac {dz}{dy}}\right|_{y(x)}\cdot \left.{\frac {dy}{dx}}\right|_{x},$$

If y = f(u) and u = g(x): $${\frac {dy}{dx}}={\frac {dy}{du}}\cdot {\frac {du}{dx}}.$$

$$\left.{\frac {dy}{dx}}\right|_{x=c}=\left.{\frac {dy}{du}}\right|_{u=g(c)}\cdot \left.{\frac {du}{dx}}\right|_{x=c}.$$

$f_{1}\circ (f_{2}\circ \cdots (f_{n-1}\circ f_{n}))\!$

$${\frac {df_{1}}{dx}}={\frac {df_{1}}{df_{2}}}{\frac {df_{2}}{df_{3}}}\cdots {\frac {df_{n}}{dx}}.$$

Chain Rule Example

$$y=e^{\sin(x^{2})}.$$

$${\begin{aligned}y&=f(u)=e^{u},\\u&=g(v)=\sin v,\\v&=h(x)=x^{2}.\end{aligned}}$$

$y=f(g(h(x)))$

$${\begin{aligned}{\frac {dy}{du}}&=f'(u)=e^{u},\\{\frac {du}{dv}}&=g'(v)=\cos v,\\{\frac {dv}{dx}}&=h'(x)=2x.\end{aligned}}$$

$${\frac {dy}{dx}}=\left.{\frac {dy}{du}}\right|_{u=g(h(a))}\cdot \left.{\frac {du}{dv}}\right|_{v=h(a)}\cdot \left.{\frac {dv}{dx}}\right|_{x=a},$$ or for short, $${\frac {dy}{dx}}={\frac {dy}{du}}\cdot {\frac {du}{dv}}\cdot {\frac {dv}{dx}}.$$ The derivative function is therefore: $${\frac {dy}{dx}}=e^{\sin(x^{2})}\cdot \cos(x^{2})\cdot 2x.$$

Chain Rule Higher Derivatives

Looks like the product rule.

f(u) and u = g(x): $${\begin{aligned}{\frac {dy}{dx}}&={\frac {dy}{du}}{\frac {du}{dx}}\\{\frac {d^{2}y}{dx^{2}}}&={\frac {d^{2}y}{du^{2}}}\left({\frac {du}{dx}}\right)^{2}+{\frac {dy}{du}}{\frac {d^{2}u}{dx^{2}}}\\{\frac {d^{3}y}{dx^{3}}}&={\frac {d^{3}y}{du^{3}}}\left({\frac {du}{dx}}\right)^{3}+3\,{\frac {d^{2}y}{du^{2}}}{\frac {du}{dx}}{\frac {d^{2}u}{dx^{2}}}+{\frac {dy}{du}}{\frac {d^{3}u}{dx^{3}}}\\{\frac {d^{4}y}{dx^{4}}}&={\frac {d^{4}y}{du^{4}}}\left({\frac {du}{dx}}\right)^{4}+6\,{\frac {d^{3}y}{du^{3}}}\left({\frac {du}{dx}}\right)^{2}{\frac {d^{2}u}{dx^{2}}}+{\frac {d^{2}y}{du^{2}}}\left(4\,{\frac {du}{dx}}{\frac {d^{3}u}{dx^{3}}}+3\,\left({\frac {d^{2}u}{dx^{2}}}\right)^{2}\right)+{\frac {dy}{du}}{\frac {d^{4}u}{dx^{4}}}.\end{aligned}}$$

Quotient Rule

The quotient rule is a consequence of the chain rule and the product rule. To see this, write the function f(x)/g(x) as the product f(x) · 1/g(x). First apply the product rule: $${\begin{aligned}{\frac {d}{dx}}\left({\frac {f(x)}{g(x)}}\right)&={\frac {d}{dx}}\left(f(x)\cdot {\frac {1}{g(x)}}\right)\\&=f'(x)\cdot {\frac {1}{g(x)}}+f(x)\cdot {\frac {d}{dx}}\left({\frac {1}{g(x)}}\right).\end{aligned}}$$

$\left({\frac {f}{g}}\right)'={\frac {f'g-g'f}{g^{2}}}\quad$

Inverse Functions

y = g(x) has an inverse function. Call its inverse function f so that we have x = f(y). $$f(g(x))=x.$$

$$f'(g(x))g'(x)=1.$$

Reciprocal rule

The derivative of $h(x)={\frac {1}{f(x)}}$for any (nonvanishing) function f is:

$h'(x)=-{\frac {f'(x)}{(f(x))^{2}}}$ wherever f is non-zero.

In Leibniz's notation, this is written

${\frac {d(1/f)}{dx}}=-{\frac {1}{f^{2}}}{\frac {df}{dx}}.$

Derivatives of exponential and logarithmic functions

${\frac {d}{dx}}\left(c^{ax}\right)={ac^{ax}\ln c},\qquad c>0$

${\frac {d}{dx}}\left(e^{ax}\right)=ae^{ax}$

${\frac {d}{dx}}\left(\log _{c}x\right)={1 \over x\ln c},\qquad c>1$

${\frac {d}{dx}}\left(\ln x\right)={1 \over x},\qquad x>0.$

${\frac {d}{dx}}\left(x^{x}\right)=x^{x}(1+\ln x).$

${\frac {d}{dx}}\left(f(x)^{g(x)}\right)=g(x)f(x)^{g(x)-1}{\frac {df}{dx}}+f(x)^{g(x)}\ln {(f(x))}{\frac {dg}{dx}},\qquad {\text{if }}f(x)>0,{\text{ and if }}{\frac {df}{dx}}{\text{ and }}{\frac {dg}{dx}}{\text{ exist.}}$

${\frac {d}{dx}}\left(f_{1}(x)^{f_{2}(x)^{\left(...\right)^{f_{n}(x)}}}\right)=\left[\sum \limits _{k=1}^{n}{\frac {\partial }{\partial x_{k}}}\left(f_{1}(x_{1})^{f_{2}(x_{2})^{\left(...\right)^{f_{n}(x_{n})}}}\right)\right]{\biggr \vert }_{x_{1}=x_{2}=...=x_{n}=x},{\text{ if }}f_{i<n}(x)>0{\text{ and }}$ ${\frac {df_{i}}{dx}}{\text{ exists. }}$

$(\ln f)'={\frac {f'}{f}}\quad$ wherever f is positive.

Derivatives of Trigonometric Functions

${\frac {d}{dx}}\sin x=\cos x$	${\frac {d}{dx}}\arcsin x={\frac {1}{\sqrt {1-x^{2}}}}$
${\frac {d}{dx}}\cos x=-\sin x$	${\frac {d}{dx}}\arccos x=-{\frac {1}{\sqrt {1-x^{2}}}}$
${\frac {d}{dx}}\tan x=\sec ^{2}x={\frac {1}{\cos ^{2}x}}=1+\tan ^{2}x$	${\frac {d}{dx}}\arctan x={\frac {1}{1+x^{2}}}$
${\frac {d}{dx}}\csc x=-\csc {x}\cot {x}$	${\frac {d}{dx}}\operatorname {arccsc} x=-{\frac {1}{|x|{\sqrt {x^{2}-1}}}}$
${\frac {d}{dx}}\sec x=\sec {x}\tan {x}$	${\frac {d}{dx}}\operatorname {arcsec} x={\frac {1}{|x|{\sqrt {x^{2}-1}}}}$
${\frac {d}{dx}}\cot x=-\csc ^{2}x=-{\frac {1}{\sin ^{2}x}}=-1-\cot ^{2}x$	${\frac {d}{dx}}\operatorname {arccot} x=-{1 \over 1+x^{2}}$

Derivatives of integrals

$F(x)=\int _{a(x)}^{b(x)}f(x,t)\,dt,$

where the functions $f(x,t)$ and ${\frac {\partial }{\partial x}}\,f(x,t)$ are both continuous in both $t$ and $x$ in some region of the $(t,x)$ plane, including $a(x)\leq t\leq b(x),$ $x_{0}\leq x\leq x_{1}$, and the functions $a(x)$ and $b(x)$ are both continuous and both have continuous derivatives for $x_{0}\leq x\leq x_{1}$. Then for $\,x_{0}\leq x\leq x_{1}$:

$F'(x)=f(x,b(x))\,b'(x)-f(x,a(x))\,a'(x)+\int _{a(x)}^{b(x)}{\frac {\partial }{\partial x}}\,f(x,t)\;dt\,.$

Integrals Rational Functions

$$\int {1 \over x}\,dx=\ln \left|x\right|+C$$

$\int a\,dx=ax+C$

$\int x^{n}\,dx={\frac {x^{n+1}}{n+1}}+C\qquad {\text{(for }}n\neq -1{\text{)}}$

$\int {1 \over x}\,dx=\ln \left|x\right|+C$

$\int {\frac {c}{ax+b}}\,dx={\frac {c}{a}}\ln \left|ax+b\right|+C$

Integrals Exponential Functions

• $\int e^{ax}\,dx={\frac {1}{a}}e^{ax}+C$
• $\int f'(x)e^{f(x)}\,dx=e^{f(x)}+C$
• $\int a^{x}\,dx={\frac {a^{x}}{\ln a}}+C$
• $\int {e^{x}\left(f\left(x\right)+f'\left(x\right)\right)\,dx}=e^{x}f\left(x\right)+C$
• $\int {e^{x}\left(f\left(x\right)-\left(-1\right)^{n}{\frac {d^{n}f\left(x\right)}{dx^{n}}}\right)\,dx}=e^{x}\sum _{k=1}^{n}{\left(-1\right)^{k-1}{\frac {d^{k-1}f\left(x\right)}{dx^{k-1}}}}+C$
  (if $n$ is a positive integer)
• $\int {e^{-x}\left(f\left(x\right)-{\frac {d^{n}f\left(x\right)}{dx^{n}}}\right)\,dx}=-e^{-x}\sum _{k=1}^{n}{\frac {d^{k-1}f\left(x\right)}{dx^{k-1}}}+C$
  (if $n$ is a positive integer)

Integrals Trigonometric functions

• $\int \sin {x}\,dx=-\cos {x}+C$
• $\int \cos {x}\,dx=\sin {x}+C$
• $\int \tan {x}\,dx=\ln {\left|\sec {x}\right|}+C=-\ln {\left|\cos {x}\right|}+C$
• $\int \arcsin {x}\,dx=x\arcsin {x}+{\sqrt {1-x^{2}}}+C,{\text{ for }}\vert x\vert \leq 1$
• $\int \arccos {x}\,dx=x\arccos {x}-{\sqrt {1-x^{2}}}+C,{\text{ for }}\vert x\vert \leq 1$
• $\int \arctan {x}\,dx=x\arctan {x}-{\frac {1}{2}}\ln {\vert 1+x^{2}\vert }+C,{\text{ for all real }}x$

Logarithms

• $\int \ln x\,dx=x\ln x-x+C=x(\ln x-1)+C$
• $\int \log _{a}x\,dx=x\log _{a}x-{\frac {x}{\ln a}}+C={\frac {x}{\ln a}}(\ln x-1)+C$

Power Rule Integration

$\int \!x^{r}\,dx={\frac {x^{r+1}}{r+1}}+C$