Matemaatika:Funktsiooni tuletis

Diferentseerimine

Valemid

$\begin{array}{c}{x}^{\prime }=1\\ (x^2{)}^{\prime }=2x\\ (x^{\alpha }{)}^{\prime }=\alpha x^{\alpha -1}\\ (\sqrt{x}{)}^{\prime }=\frac{1}{2\sqrt{x}}\\ (\frac{1}{x}{)}^{\prime }=-\frac{1}{x^2}\\ (e^x{)}^{\prime }=e^x\\ (ln(x){)}^{\prime }=\frac{1}{x}\\ (lo{g}_{a}x{)}^{\prime }=\frac{1}{xln(a)}\end{array}$$\begin{array}{c}(a{)}^{\prime }=0\\ (a\ast u{)}^{\prime }=a\ast (u{)}^{\prime }\\ (u+v{)}^{\prime }=u^{\prime }+v^{\prime }\\ (u-v{)}^{\prime }=u^{\prime }-v^{\prime }\\ (u\ast v{)}^{\prime }=u^{\prime }\ast v+u\ast v^{\prime }\\ (\frac{u}{v}{)}^{\prime }=\frac{u^{\prime }\ast v-u\ast v^{\prime }}{v^2}\\ (a^x{)}^{\prime }=a^{x}ln(a)\end{array}$$\begin{array}{c}(sin(x){)}^{\prime }=cos(x)\\ (cos(x){)}^{\prime }=-sin(x)\\ (tan(x){)}^{\prime }=\frac{1}{co{s}^2(x)}\\ (cot(x){)}^{\prime }=\frac{1}{si{n}^2(x)}\\ (arcsin(x){)}^{\prime }=\frac{1}{\sqrt{1-x^2}}\\ (arccos(x){)}^{\prime }=-\frac{1}{\sqrt{1-x^2}}\\ (arctan(x){)}^{\prime }=\frac{1}{1+x^2}\\ (arccottan(x){)}^{\prime }=-\frac{1}{1+x^2}\end{array}$

Meeldetuletuseks

Tähemärk a on konstant, mis on kindla ühe arvväärtusega element. Arv e=~2,71828182845905...
$\begin{array}{c}ln(a\ast b)=ln(a)+ln(b)\\ ln(\frac{a}{b})=ln(a)-ln(b)\\ ln(a^b)=b\ast ln(a)\end{array}$

$e=\underset{x\to \mathrm{\infty }}{\lim }(1+\frac{1}{x}{)}^x$

Logaritmiline diferentseerimine

• korrutan mõlemad pooled naturaallogaritmiga läbi
• diferentseerin x järgi
• avaldan tuletise
• asendan y tema algväärtusega

Ülesanne 1

$\begin{array}{c}y=x^x\end{array}$,leida y' $\begin{array}{c}y=x^x|ln\\ ln(y)=x\ast ln(x)|\frac{d}{dx}\\ \frac{1}{y}\ast y^{\prime }=1\ast ln(x)+x\ast \frac{1}{x}\\ y^{\prime }=y(ln(x)+1)=x^x(ln(x)+1)\end{array}$

Ülesanne 2

$\begin{array}{c}y=x^{x^x}\end{array}$,leida y' $\begin{array}{c}y=x^{x^x}|ln\\ ln(y)=x^x\ast ln(x)|ln\\ ln(ln(y))=x\ast ln(x)+ln(ln(x))|\frac{d}{dx}\\ \frac{1}{ln(y)}\ast \frac{1}{y}\ast y^{\prime }=1\ast ln(x)+x\ast \frac{1}{x}+\frac{1}{ln(x)}\ast \frac{1}{x}\\ y^{\prime }=y\ast ln(y)\ast (ln(x)+1+\frac{1}{x\ast ln(x)})\\ y^{\prime }=x^{x^x}\ast x^x\ast ln(x)\ast (ln(x)+1+\frac{1}{x\ast ln(x)})\end{array}$