Definiție
$$
\begin{aligned}
&\text{Fie funcțiile }\quad u:A\to B,\qquad f:B\to \mathbb{R}.\\[6pt]
&\text{Dacă }\; u \text{ este derivabilă în } x_0\in A,\quad
f \text{ este derivabilă în } u(x_0)\in B,\\[6pt]
&\text{atunci }\; (f\circ u):A\to \mathbb{R}
\text{ este derivabilă în punctul } x_0
\text{ și are loc relația:}\\[6pt]
&(f\circ u)'(x_0)=f'(u(x_0))\cdot u'(x_0).
\end{aligned}
$$
\[
(f\circ u)’=(f’\circ u)\cdot u’
\]
Formule
\[
(u^n)’=n\cdot u^{n-1}\cdot u’,\quad n\in\mathbb{N}^*
\]
\[
(u^r)’=r\cdot u^{r-1}\cdot u’,\quad r\in\mathbb{R}
\]
\[
(\sqrt{u})’=\frac{1}{2\sqrt{u}}\cdot u’
\]
\[
(\sqrt[n]{u})’=\frac{1}{n\sqrt[n]{u^{\,n-1}}}\cdot u’
\]
\[
(\ln u)’=\frac{1}{u}\cdot u’
\]
\[
(\log_a u)’=\frac{1}{u\ln a}\cdot u’
\]
\[
(e^u)’=e^u\cdot u’
\]
\[
(a^u)’=a^u\cdot \ln a\cdot u’
\]
\[
(\sin u)’=\cos u\cdot u’
\]
\[
(\cos u)’=-\sin u\cdot u’
\]
\[
(\mathrm{tg}\,u)’=\frac{1}{\cos^2 u}\cdot u’
\]
\[
(\mathrm{ctg}\,u)’=-\frac{1}{\sin^2 u}\cdot u’
\]
Exerciții rezolvate
$$
\begin{aligned}
&1.\quad
\left[(2x-5)^3\right]’
=3(2x-5)^2\cdot(2x-5)’
=3(2x-5)^2\cdot 2
=6(2x-5)^2
\\[10pt]
&2.\quad
\left[(5-3x)^{10}\right]’
=10(5-3x)^9\cdot(5-3x)’
=10(5-3x)^9\cdot(-3)
=\\[10pt]
&\phantom{2.\quad}
=-30(5-3x)^9
\\[10pt]
&3.\quad
\left[(x^2-3x-5)^{201}\right]’
=201(x^2-3x-5)^{200}\cdot(x^2-3x-5)’
=\\[10pt]
&\phantom{3.\quad}
=201(x^2-3x-5)^{200}\cdot(2x-3)
\end{aligned}
$$
$$
\begin{aligned}
&4.\quad
\left(\sqrt{5x-7}\right)’
=\frac{1}{2\sqrt{5x-7}}\cdot(5x-7)’
=\frac{1}{2\sqrt{5x-7}}\cdot 5
=\frac{5}{2\sqrt{5x-7}}
\\[10pt]
&5.\quad
\left(\sqrt{1-x+2x^2}\right)’
=\frac{1}{2\sqrt{1-x+2x^2}}\cdot(1-x+2x^2)’
=\\[10pt]
&\phantom{5.\quad}
=\frac{1}{2\sqrt{1-x+2x^2}}\cdot(-1+4x)
=\frac{4x-1}{2\sqrt{1-x+2x^2}}
\\[10pt]
&6.\quad
\left(\sqrt[3]{3x-7}\right)’
=\frac{1}{3\sqrt[3]{(3x-7)^2}}\cdot(3x-7)’
=\frac{1}{3\sqrt[3]{(3x-7)^2}}\cdot 3
=\\[10pt]
&\phantom{6.\quad}
=\frac{1}{\sqrt[3]{(3x-7)^2}}
\\[10pt]
&7.\quad
\left(\sqrt[3]{4x^2-3x+1}\right)’
=\frac{1}{3\sqrt[3]{(4x^2-3x+1)^2}}\cdot(4x^2-3x+1)’
=\\[10pt]
&\phantom{7.\quad}
=\frac{1}{3\sqrt[3]{(4x^2-3x+1)^2}}\cdot(8x-3)
=\frac{8x-3}{3\sqrt[3]{(4x^2-3x+1)^2}}
\end{aligned}
$$
$$
\begin{aligned}
&8.\quad
\left(e^{2x}\right)’
=e^{2x}\cdot(2x)’
=e^{2x}\cdot 2
=2e^{2x}
\\[10pt]
&9.\quad
\left(e^{4-7x}\right)’
=e^{4-7x}\cdot(4-7x)’
=e^{4-7x}\cdot(-7)
=-7e^{4-7x}
\\[10pt]
&10.\quad
\left(e^{2x^2-3x+1}\right)’
=e^{2x^2-3x+1}\cdot(2x^2-3x+1)’
=\\[10pt]
&\phantom{10.\quad}
=e^{2x^2-3x+1}\cdot(4x-3)
=(4x-3)e^{2x^2-3x+1}
\end{aligned}
$$
$$
\begin{aligned}
&11.\quad
\left(2^{3x-1}\right)’
=2^{3x-1}\cdot \ln 2\cdot(3x-1)’
=2^{3x-1}\cdot \ln 2\cdot 3
=\\[10pt]
&\phantom{11.\quad}
=3\cdot 2^{3x-1}\ln 2
\\[14pt]
&12.\quad
\left[\left(\frac{1}{3}\right)^{1-x^2}\right]’
=\left(\frac{1}{3}\right)^{1-x^2}\cdot \ln\frac{1}{3}\cdot(1-x^2)’
=\\[10pt]
&\phantom{12.\quad}
=\left(\frac{1}{3}\right)^{1-x^2}\cdot \ln\frac{1}{3}\cdot(-2x)
=\\[10pt]
&\phantom{12.\quad}
=-2x\left(\frac{1}{3}\right)^{1-x^2}\ln\frac{1}{3}
\\[14pt]
&13.\quad
\left(\sqrt{2}^{\,x^2-x-3}\right)’
=\left(\sqrt{2}\right)^{x^2-x-3}\cdot \ln\sqrt{2}\cdot(x^2-x-3)’
=\\[10pt]
&\phantom{13.\quad}
=\left(\sqrt{2}\right)^{x^2-x-3}\cdot \ln\sqrt{2}\cdot(2x-1)
=\\[10pt]
&\phantom{13.\quad}
=(2x-1)\left(\sqrt{2}\right)^{x^2-x-3}\ln\sqrt{2}
\end{aligned}
$$
$$
\begin{aligned}
&14.\quad
\left[\ln(10x-9)\right]’
=\frac{1}{10x-9}\cdot(10x-9)’
=\frac{10}{10x-9}
\\[14pt]
&15.\quad
\left[\ln(1-x+3x^2)\right]’
=\frac{1}{1-x+3x^2}\cdot(1-x+3x^2)’
=\frac{-1+6x}{1-x+3x^2}
\\[14pt]
&16.\quad
\left[\log_5(1-3x)\right]’
=\frac{1}{(1-3x)\ln 5}\cdot(1-3x)’
=-\frac{3}{(1-3x)\ln 5}
\\[14pt]
&17.\quad
\left[\log_{\frac{1}{3}}(x^2-3x+7)\right]’
=\frac{1}{(x^2-3x+7)\ln\frac{1}{3}}\cdot(x^2-3x+7)’
=\\[10pt]
&\phantom{17.\quad}
=\frac{2x-3}{(x^2-3x+7)\ln\frac{1}{3}}
\end{aligned}
$$
$$
\begin{aligned}
&18.\quad
\left[\sin(4x-1)\right]’
=\cos(4x-1)\cdot(4x-1)’
=4\cos(4x-1)
\\[14pt]
&19.\quad
\left[\cos(1-5x)\right]’
=-\sin(1-5x)\cdot(1-5x)’
=5\sin(1-5x)
\\[14pt]
&20.\quad
\left[\mathrm{tg}(6x+5)\right]’
=\frac{1}{\cos^2(6x+5)}\cdot(6x+5)’
=\frac{6}{\cos^2(6x+5)}
\\[14pt]
&21.\quad
\left[\mathrm{ctg}(5-8x)\right]’
=-\frac{1}{\sin^2(5-8x)}\cdot(5-8x)’
=\frac{8}{\sin^2(5-8x)}
\end{aligned}
$$
$$
\begin{aligned}
&22.\quad
\left[(2x-1)^2\cdot e^{x+3}\right]’
=\left[(2x-1)^2\right]’\cdot e^{x+3}
+(2x-1)^2\cdot\left(e^{x+3}\right)’
=\\[10pt]
&\phantom{22.\quad}
=2(2x-1)\cdot2\cdot e^{x+3}
+(2x-1)^2\cdot e^{x+3}
\\[14pt]
&23.\quad
\left[\sin(2x+3)\cdot e^{2x-1}\right]’
=\cos(2x+3)\cdot(2x+3)’\cdot e^{2x-1}
+\\[10pt]
&\phantom{23.\quad}
+\sin(2x+3)\cdot e^{2x-1}\cdot(2x-1)’
=\\[10pt]
&\phantom{23.\quad}
=2\cos(2x+3)e^{2x-1}
+2\sin(2x+3)e^{2x-1}
\\[14pt]
&24.\quad
\left[e^{x^2-x+1}\cdot(x-1)^2\right]’
=e^{x^2-x+1}\cdot(x^2-x+1)’\cdot(x-1)^2
+\\[10pt]
&\phantom{24.\quad}
+e^{x^2-x+1}\cdot\left[(x-1)^2\right]’
=\\[10pt]
&\phantom{24.\quad}
=e^{x^2-x+1}(2x-1)(x-1)^2
+e^{x^2-x+1}\cdot2(x-1)
\end{aligned}
$$
$$
\begin{aligned}
&26.\quad
\left[\ln\left(x+\sqrt{x}\right)\right]’
=\frac{1}{x+\sqrt{x}}\cdot\left(x+\sqrt{x}\right)’
=\\[10pt]
&\phantom{26.\quad}
=\frac{1}{x+\sqrt{x}}\cdot\left(1+\frac{1}{2\sqrt{x}}\right)
\\[14pt]
&27.\quad
\left[\ln^3\left(x^2-x+3\right)\right]’
=3\ln^2\left(x^2-x+3\right)\cdot
\left[\ln\left(x^2-x+3\right)\right]’
=\\[10pt]
&\phantom{27.\quad}
=3\ln^2\left(x^2-x+3\right)\cdot
\frac{2x-1}{x^2-x+3}
=\\[10pt]
&\phantom{27.\quad}
=\frac{3(2x-1)\ln^2\left(x^2-x+3\right)}{x^2-x+3}
\end{aligned}
$$
$$
\begin{aligned}
&28.\quad
\left[\sin^2(4x-7)\right]’
=2\sin(4x-7)\cdot\left[\sin(4x-7)\right]’
=\\[10pt]
&\phantom{28.\quad}
=2\sin(4x-7)\cdot\cos(4x-7)\cdot(4x-7)’
=\\[10pt]
&\phantom{28.\quad}
=8\sin(4x-7)\cos(4x-7)
\\[14pt]
&29.\quad
\left[\cos^3(x+2)\right]’
=3\cos^2(x+2)\cdot\left[\cos(x+2)\right]’
=\\[10pt]
&\phantom{29.\quad}
=3\cos^2(x+2)\cdot\left[-\sin(x+2)\right]\cdot(x+2)’
=\\[10pt]
&\phantom{29.\quad}
=-3\cos^2(x+2)\sin(x+2)
\\[14pt]
&30.\quad
\left[\left(\frac{1-e^x}{1+e^x}\right)^3\right]’
=3\left(\frac{1-e^x}{1+e^x}\right)^2
\cdot
\left(\frac{1-e^x}{1+e^x}\right)’
=\\[10pt]
&\phantom{30.\quad}
=3\left(\frac{1-e^x}{1+e^x}\right)^2
\cdot
\frac{-e^x(1+e^x)-(1-e^x)e^x}{(1+e^x)^2}
=\\[10pt]
&\phantom{30.\quad}
=3\left(\frac{1-e^x}{1+e^x}\right)^2
\cdot
\frac{-2e^x}{(1+e^x)^2}
=\\[10pt]
&\phantom{30.\quad}
=-\frac{6e^x(1-e^x)^2}{(1+e^x)^4}
\\[14pt]
&31.\quad
\left[\frac{1}{(x^2+1)^4}\right]’
=\left[(x^2+1)^{-4}\right]’
=-4(x^2+1)^{-5}\cdot(x^2+1)’
=\\[10pt]
&\phantom{31.\quad}
=-4(x^2+1)^{-5}\cdot2x
=-8x(x^2+1)^{-5}
=\\[10pt]
&\phantom{31.\quad}
=-\frac{8x}{(x^2+1)^5}
\end{aligned}
$$
\[
(u^v)’=v\cdot u^{v-1}\cdot u’ + u^v\cdot \ln u \cdot v’
\]
$$
\begin{aligned}
&1.\quad \text{Să se calculeze derivata funcției } f(x)=x^x.
\\[14pt]
&\textbf{Metoda 1. Folosind formula}
\\[10pt]
&(x^x)’
=x\cdot x^{x-1}\cdot 1
+x^x\cdot\ln x\cdot 1
=\\[10pt]
&\phantom{(x^x)’}
=x^x+x^x\ln x
=x^x(1+\ln x)
\\[16pt]
&\textbf{Metoda 2. Folosind scrierea cu } e
\\[10pt]
&\color{red}{f^g=e^{g\ln f}}
\\[10pt]
&x^x=e^{x\ln x}
\\[10pt]
&(x^x)’
=\left(e^{x\ln x}\right)’
=e^{x\ln x}\cdot(x\ln x)’
=\\[10pt]
&\phantom{(x^x)’}
=x^x\cdot\left(x’\ln x+x\cdot(\ln x)’\right)
=x^x\left(\ln x+x\cdot\frac{1}{x}\right)
=\\[10pt]
&\phantom{(x^x)’}
=x^x(1+\ln x)
\end{aligned}
$$
$$
\begin{aligned}
&2.\quad \text{Să se calculeze derivata funcției } f(x)=x^{\sqrt{x}}.
\\[14pt]
&\left(x^{\sqrt{x}}\right)’
=\sqrt{x}\cdot x^{\sqrt{x}-1}\cdot 1
+x^{\sqrt{x}}\cdot\ln x\cdot(\sqrt{x})’
=\\[10pt]
&\phantom{\left(x^{\sqrt{x}}\right)’}
=\sqrt{x}\cdot x^{\sqrt{x}-1}
+x^{\sqrt{x}}\cdot\ln x\cdot\frac{1}{2\sqrt{x}}
\\[18pt]
&3.\quad \text{Să se calculeze derivata funcției } f(x)=(x+1)^{x+2}.
\\[14pt]
&\left[(x+1)^{x+2}\right]’
=(x+2)(x+1)^{x+1}\cdot(x+1)’
+
(x+1)^{x+2}\cdot\ln(x+1)\cdot(x+2)’
=\\[10pt]
&\phantom{\left[(x+1)^{x+2}\right]’}
=(x+2)(x+1)^{x+1}
+(x+1)^{x+2}\ln(x+1)
\end{aligned}
$$
