User:JaviPrieto/Derivatives

(Click for larger image) At each point, the derivative of $\scriptstyle f(x)=1+x\sin x^{2}$ is the slope of a line that is tangent to the curve. The line is always tangent to the blue curve; its slope is the derivative. Note derivative is positive where green, negative where red, and zero where black.

$m={{\mbox{change in }}y \over {\mbox{change in }}x}={\Delta y \over {\Delta x}}$

${\frac {dy}{dx}}\,\!$

$m={\frac {\Delta f(x)}{\Delta x}}={\frac {f(x+h)-f(x)}{h}}.$

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

$\lim _{h\to 0}{f(a+h)-f(a)-f'(a)\cdot h \over h}=0,$

$f(a+h)\approx f(a)+f'(a)h$

$Q(h)={\frac {f(a+h)-f(a)}{h}}.$

Q(h) is the slope of the secant line between (a, ç'(a)) and (a + h, ç'(a + h)). If ç' is a continuous function, meaning that its graph is an unbroken curve with no gaps, then Q is a continuous function away from the point h = 0. If the limit $\textstyle \lim _{h\to 0}Q(h)$ exists, meaning that there is a way of choosing a value for Q(0) that makes the graph of Q a continuous function, then the function ç' is differentiable at the point a, and its derivative at a equals Q(0).

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

$\lim _{h\to 0}{6+h}=6+0=6.$

${\begin{aligned}1&{}\mapsto 2,\\2&{}\mapsto 4,\\3&{}\mapsto 6.\end{aligned}}$

${\begin{aligned}D(x\mapsto 1)&=(x\mapsto 0),\\D(x\mapsto x)&=(x\mapsto 1),\\D(x\mapsto x^{2})&=(x\mapsto 2\cdot x).\end{aligned}}$

$x\mapsto x^{2},$

$x\mapsto 2x,$

$f(x)={\begin{cases}x^{2},&{\mbox{if }}x\geq 0\\-x^{2},&{\mbox{if }}x\leq 0.\end{cases}}$

$f'(x)={\begin{cases}2x,&{\mbox{if }}x\geq 0\\-2x,&{\mbox{if }}x\leq 0.\end{cases}}$

$f(x+h)\approx f(x)+f'(x)h+{\tfrac {1}{2}}f''(x)h^{2}$

$\lim _{h\to 0}{\frac {f(x+h)-f(x)-f'(x)h-{\frac {1}{2}}f''(x)h^{2}}{h^{2}}}=0.$

${\frac {dy}{dx}},\quad {\frac {df}{dx}}(x),\;\;\mathrm {or} \;\;{\frac {d}{dx}}f(x),$

${\frac {d^{n}y}{dx^{n}}},\quad {\frac {d^{n}f}{dx^{n}}}(x),\;\;\mathrm {or} \;\;{\frac {d^{n}}{dx^{n}}}f(x)$

${\frac {d^{2}y}{dx^{2}}}={\frac {d}{dx}}\left({\frac {dy}{dx}}\right).$

$\left.{\frac {dy}{dx}}\right|_{x=a}={\frac {dy}{dx}}(a).$

${\frac {dy}{dx}}={\frac {dy}{du}}\cdot {\frac {du}{dx}}.$

$(f')'=f''\,$   and   $(f'')'=f'''.\,$

$f^{\mathrm {iv} }\,\!$   or   $f^{(4)}.\,\!$

${\dot {y}}$   and   ${\ddot {y}}$

$D_{x}y\,$   or   $D_{x}f(x)\,$,

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

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

wherever this function is defined. For example, if $f(x)=x^{1/4}$, then

$f'(x)=(1/4)x^{-3/4},\,$

${\frac {d}{dx}}e^{x}=e^{x}$

${\frac {d}{dx}}a^{x}=\ln(a)a^{x}$

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

${\frac {d}{dx}}\log _{a}(x)={\frac {1}{x\ln(a)}}$

${\frac {d}{dx}}\sin(x)=\cos(x).$

${\frac {d}{dx}}\cos(x)=-\sin(x).$

${\frac {d}{dx}}\tan(x)=\sec ^{2}(x)={\frac {1}{\cos ^{2}(x)}}=1+\tan ^{2}(x).$

${\frac {d}{dx}}\arcsin(x)={\frac {1}{\sqrt {1-x^{2}}}}.$

${\frac {d}{dx}}\arccos(x)=-{\frac {1}{\sqrt {1-x^{2}}}}.$

${\frac {d}{dx}}\arctan(x)={\frac {1}{1+x^{2}}}.$

$f'=0\,$

$(af+bg)'=af'+bg'\,$ for all functions ƒ and g and all real numbers a and b.

$(fg)'=f'g+fg'\,$ for all functions ƒ and g.

$\left({\frac {f}{g}}\right)'={\frac {f'g-fg'}{g^{2}}}$ for all functions ƒ and g where g ? 0.

• Chain rule: If $f(x)=h(g(x))$, then

$f'(x)=h'(g(x))\cdot g'(x).\,$

$f(x)=x^{4}+\sin(x^{2})-\ln(x)e^{x}+7\,$

${\begin{aligned}f'(x)&=4x^{(4-1)}+{\frac {d\left(x^{2}\right)}{dx}}\cos(x^{2})-{\frac {d\left(\ln {x}\right)}{dx}}e^{x}-\ln {x}{\frac {d\left(e^{x}\right)}{dx}}+0\\&=4x^{3}+2x\cos(x^{2})-{\frac {1}{x}}e^{x}-\ln(x)e^{x}.\end{aligned}}$

$\mathbf {y} '(t)=(y'_{1}(t),\ldots ,y'_{n}(t)).$

$\mathbf {y} '(t)=\lim _{h\to 0}{\frac {\mathbf {y} (t+h)-\mathbf {y} (t)}{h}},$

$y'_{1}(t)\mathbf {e} _{1}+\cdots +y'_{n}(t)\mathbf {e} _{n}$

$f(x,y)=x^{2}+xy+y^{2}.\,$

$f(x,y)=f_{x}(y)=x^{2}+xy+y^{2}.\,$

$x\mapsto f_{x},\,$

$f_{x}(y)=x^{2}+xy+y^{2}.\,$

$f_{a}(y)=a^{2}+ay+y^{2}.\,$

$f_{a}'(y)=a+2y.\,$

${\frac {\partial f}{\partial y}}(x,y)=x+2y.$

${\frac {\partial f}{\partial x_{i}}}(a_{1},\ldots ,a_{n})=\lim _{h\to 0}{\frac {f(a_{1},\ldots ,a_{i}+h,\ldots ,a_{n})-f(a_{1},\ldots ,a_{n})}{h}}.$

$f_{a_{1},\ldots ,a_{i-1},a_{i+1},\ldots ,a_{n}}(x_{i})=f(a_{1},\ldots ,a_{i-1},x_{i},a_{i+1},\ldots ,a_{n})$

${\frac {df_{a_{1},\ldots ,a_{i-1},a_{i+1},\ldots ,a_{n}}}{dx_{i}}}(a_{i})={\frac {\partial f}{\partial x_{i}}}(a_{1},\ldots ,a_{n}).$

$\nabla f(a)=\left({\frac {\partial f}{\partial x_{1}}}(a),\ldots ,{\frac {\partial f}{\partial x_{n}}}(a)\right).$

$\mathbf {v} =(v_{1},\ldots ,v_{n}).$

$D_{\mathbf {v} }{f}(\mathbf {x} )=\lim _{h\rightarrow 0}{\frac {f(\mathbf {x} +h\mathbf {v} )-f(\mathbf {x} )}{h}}.$

${\frac {f(\mathbf {x} +(k/\lambda )(\lambda \mathbf {u} ))-f(\mathbf {x} )}{k/\lambda }}=\lambda \cdot {\frac {f(\mathbf {x} +k\mathbf {u} )-f(\mathbf {x} )}{k}}.$

$D_{\mathbf {v} }{f}({\boldsymbol {x}})=\sum _{j=1}^{n}v_{j}{\frac {\partial f}{\partial x_{j}}}.$

$f(\mathbf {a} +\mathbf {v} )\approx f(\mathbf {a} )+f'(\mathbf {a} )\mathbf {v} .$

$f(\mathbf {a} +\mathbf {v} )-f(\mathbf {a} )\approx f'(\mathbf {a} )\mathbf {v} .$

$f(\mathbf {a} +\mathbf {v} +\mathbf {w} )-f(\mathbf {a} +\mathbf {v} )-f(\mathbf {a} +\mathbf {w} )+f(\mathbf {a} )\approx f'(\mathbf {a} +\mathbf {v} )\mathbf {w} -f'(\mathbf {a} )\mathbf {w} .$

${\begin{aligned}0&\approx f(\mathbf {a} +\mathbf {v} +\mathbf {w} )-f(\mathbf {a} +\mathbf {v} )-f(\mathbf {a} +\mathbf {w} )+f(\mathbf {a} )\\&=(f(\mathbf {a} +\mathbf {v} +\mathbf {w} )-f(\mathbf {a} ))-(f(\mathbf {a} +\mathbf {v} )-f(\mathbf {a} ))-(f(\mathbf {a} +\mathbf {w} )-f(\mathbf {a} ))\\&\approx f'(\mathbf {a} )(\mathbf {v} +\mathbf {w} )-f'(\mathbf {a} )\mathbf {v} -f'(\mathbf {a} )\mathbf {w} .\end{aligned}}$

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

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

$\lim _{\lVert \mathbf {h} \rVert \to 0}{\frac {\lVert f(\mathbf {a} +\mathbf {h} )-f(\mathbf {a} )-f'(\mathbf {a} )\mathbf {h} \rVert }{\lVert \mathbf {h} \rVert }}=0.$

$f'(\mathbf {a} )=\operatorname {Jac} _{\mathbf {a} }=\left({\frac {\partial f_{i}}{\partial x_{j}}}\right)_{ij}.$

$f(a+h)\approx f(a)+f'(a)h.$

Up to changing variables, this is the statement that the function $x\mapsto f(a)+f'(a)(x-a)$ is the best linear approximation to ç' at a.