\((c)’=0\)
Exemple:
\(1.\quad (2)’=0\)
\(2.\quad (-5)’=0\)
\(3.\quad \left(\frac{7}{3}\right)’=0\)
\((x^n)’=n\cdot x^{n-1}\)
Exemple:
\(1.\quad (x^2)’=2x\)
\(2.\quad (x^3)’=3x^2\)
\(3.\quad (x^5)’=5x^4\)
\(\left(\sqrt[n]{x}\right)’=\frac{1}{n\sqrt[n]{x^{\,n-1}}}\)
Exemple:
\(1.\quad (\sqrt{x})’=\frac{1}{2\sqrt{x}}\)
\(2.\quad \left(\sqrt[3]{x}\right)’=\frac{1}{3\sqrt[3]{x^2}}\)
\(3.\quad \left(\sqrt[4]{x}\right)’=\frac{1}{4\sqrt[4]{x^3}}\)
\((a^x)’=a^x\ln a,\quad a>0,\ a\neq 1;\qquad (e^x)’=e^x\)
Exemple:
\(1.\quad (2^x)’=2^x\ln 2\)
\(2.\quad \left(\left(\frac{1}{3}\right)^x\right)’=\left(\frac{1}{3}\right)^x\ln\frac{1}{3}\)
\(3.\quad (e^x)’=e^x\)
\((\log_a x)’=\frac{1}{x\ln a},\quad a>0,\ a\neq 1,\ x>0;\qquad (\ln x)’=\frac{1}{x},\ x>0\)
Exemple:
\(1.\quad (\log_2 x)’=\frac{1}{x\ln 2}\)
\(2.\quad \left(\log_{\frac{1}{3}} x\right)’=\frac{1}{x\ln \frac{1}{3}}\)
\(3.\quad (\ln x)’=\frac{1}{x}\)
\((\sin x)’=\cos x;\qquad (\cos x)’=-\sin x\)
Exemple:
\(1.\quad (\sin x)’=\cos x\)
\(2.\quad (\cos x)’=-\sin x\)
\(3.\quad (-\cos x)’=\sin x\)
\((\mathrm{tg}\,x)’=\frac{1}{\cos^2 x};\qquad (\mathrm{ctg}\,x)’=-\frac{1}{\sin^2 x}\)
Exemple:
\(1.\quad (-\mathrm{tg}\,x)’=-\frac{1}{\cos^2 x}\)
\(2.\quad (-\mathrm{ctg}\,x)’=\frac{1}{\sin^2 x}\)
\((f\pm g)’=f’\pm g’\)
Exemple:
\(1.\quad (2x+1)’=(2x)’+(1)’=2+0=2\)
\(2.\quad (x^2-3x+7)’=(x^2)’-(3x)’+(7)’=2x-3+0=2x-3\)
\(3.\quad (x^3+4x^2-5)’=(x^3)’+(4x^2)’-(5)’=3x^2+8x-0=3x^2+8x\)
\(4.\quad (5x^4-2x^3+x-9)’=(5x^4)’-(2x^3)’+(x)’-(9)’=20x^3-6x^2+1-0=20x^3-6x^2+1\)
\(5.\quad (-3x^5+7x^2-4x+6)’=(-3x^5)’+(7x^2)’-(4x)’+(6)’=-15x^4+14x-4+0=-15x^4+14x-4\)
\(6.\quad (x^6-2x^4+3x^3-x+8)’=(x^6)’-(2x^4)’+(3x^3)’-(x)’+(8)’=6x^5-8x^3+9x^2-1+0=6x^5-8x^3+9x^2-1\)
\(7.\quad (x+\sqrt{x})’=(x)’+(\sqrt{x})’=1+\frac{1}{2\sqrt{x}}\)
\(8.\quad (2^x-2x+3)’=(2^x)’-(2x)’+(3)’=2^x\ln 2-2+0=2^x\ln 2-2\)
\(9.\quad (x^2+\ln x)’=(x^2)’+(\ln x)’=2x+\frac{1}{x}\)
\(10.\quad (e^x-3x^2+5)’=(e^x)’-(3x^2)’+(5)’=e^x-6x+0=e^x-6x\)
\(
11.\quad (2\sin x-3\cos x+4)’=(2\sin x)’-(3\cos x)’+(4)’=2\cos x+3\sin x
\)
\( 12.\quad \left(-\frac{1}{2}\cos x+\frac{2}{3}\sin x-3x+5\right)’=\left(-\frac{1}{2}\cos x\right)’+\left(\frac{2}{3}\sin x\right)’-(3x)’+(5)’=\frac{1}{2}\sin x+\frac{2}{3}\cos x-3 \)
\( 13.\quad (\tan x-\cot x+3)’=(\tan x)’-(\cot x)’+(3)’=\frac{1}{\cos^2 x}+\frac{1}{\sin^2 x} \)
\( 14.\quad \left(3\cot x-5\tan x-4x+\frac{1}{3}\right)’=(3\cot x)’-(5\tan x)’-(4x)’+\left(\frac{1}{3}\right)’=-\frac{3}{\sin^2 x}-\frac{5}{\cos^2 x}-4 \)
\( 12.\quad \left(-\frac{1}{2}\cos x+\frac{2}{3}\sin x-3x+5\right)’=\left(-\frac{1}{2}\cos x\right)’+\left(\frac{2}{3}\sin x\right)’-(3x)’+(5)’=\frac{1}{2}\sin x+\frac{2}{3}\cos x-3 \)
\( 13.\quad (\tan x-\cot x+3)’=(\tan x)’-(\cot x)’+(3)’=\frac{1}{\cos^2 x}+\frac{1}{\sin^2 x} \)
\( 14.\quad \left(3\cot x-5\tan x-4x+\frac{1}{3}\right)’=(3\cot x)’-(5\tan x)’-(4x)’+\left(\frac{1}{3}\right)’=-\frac{3}{\sin^2 x}-\frac{5}{\cos^2 x}-4 \)
\[
(f\cdot g)’=f’\cdot g+f\cdot g’
\]
\(
1.\quad \left(x^2(x+1)\right)’=(x^2)'(x+1)+x^2(x+1)’=
\)
\( \phantom{1.\quad}=2x(x+1)+x^2=2x^2+2x+x^2=3x^2+2x \)
\( 2.\quad \left((2x+5)(1-3x)\right)’=(2x+5)'(1-3x)+(2x+5)(1-3x)’= \)
\( \phantom{2.\quad}=2(1-3x)+(2x+5)(-3)=2-6x-6x-15=-12x-13 \)
\( 3.\quad \left((x^2+1)(3x-5)\right)’=(x^2+1)'(3x-5)+(x^2+1)(3x-5)’= \)
\( \phantom{3.\quad}=2x(3x-5)+3(x^2+1)=6x^2-10x+3x^2+3=9x^2-10x+3 \)
\( 4.\quad \left((x^2-x+2)(x+3)\right)’=(x^2-x+2)'(x+3)+(x^2-x+2)(x+3)’= \)
\( \phantom{4.\quad}=(2x-1)(x+3)+(x^2-x+2)=2x^2+6x-x-3+x^2-x+2 \)
\( \phantom{4.\quad}=3x^2+4x-1 \)
\( 5.\quad \left((2x-7)(2x^2-3x+1)\right)’=(2x-7)'(2x^2-3x+1)+(2x-7)(2x^2-3x+1)’= \)
\( \phantom{5.\quad}=2(2x^2-3x+1)+(2x-7)(4x-3) \)
\( \phantom{5.\quad}=4x^2-6x+2+8x^2-6x-28x+21=12x^2-40x+23 \)
\( \phantom{1.\quad}=2x(x+1)+x^2=2x^2+2x+x^2=3x^2+2x \)
\( 2.\quad \left((2x+5)(1-3x)\right)’=(2x+5)'(1-3x)+(2x+5)(1-3x)’= \)
\( \phantom{2.\quad}=2(1-3x)+(2x+5)(-3)=2-6x-6x-15=-12x-13 \)
\( 3.\quad \left((x^2+1)(3x-5)\right)’=(x^2+1)'(3x-5)+(x^2+1)(3x-5)’= \)
\( \phantom{3.\quad}=2x(3x-5)+3(x^2+1)=6x^2-10x+3x^2+3=9x^2-10x+3 \)
\( 4.\quad \left((x^2-x+2)(x+3)\right)’=(x^2-x+2)'(x+3)+(x^2-x+2)(x+3)’= \)
\( \phantom{4.\quad}=(2x-1)(x+3)+(x^2-x+2)=2x^2+6x-x-3+x^2-x+2 \)
\( \phantom{4.\quad}=3x^2+4x-1 \)
\( 5.\quad \left((2x-7)(2x^2-3x+1)\right)’=(2x-7)'(2x^2-3x+1)+(2x-7)(2x^2-3x+1)’= \)
\( \phantom{5.\quad}=2(2x^2-3x+1)+(2x-7)(4x-3) \)
\( \phantom{5.\quad}=4x^2-6x+2+8x^2-6x-28x+21=12x^2-40x+23 \)
\(
6.\quad (e^x(x+1))’=(e^x)'(x+1)+e^x(x+1)’=
\)
\( \phantom{6.\quad}=e^x(x+1)+e^x\cdot 1=e^x(x+1+1)=e^x(x+2) \)
\( 7.\quad ((2x-5)e^x)’=(2x-5)’e^x+(2x-5)(e^x)’= \)
\( \phantom{7.\quad}=2e^x+(2x-5)e^x=e^x(2+2x-5)=e^x(2x-3) \)
\( 8.\quad (e^x(3x^2-2x-5))’=(e^x)'(3x^2-2x-5)+e^x(3x^2-2x-5)’= \)
\( \phantom{8.\quad}=e^x(3x^2-2x-5)+e^x(6x-2)=e^x(3x^2-2x-5+6x-2) \)
\( \phantom{8.\quad}=e^x(3x^2+4x-7) \)
\( 9.\quad (e^x(e^x-1))’=(e^x)'(e^x-1)+e^x(e^x-1)’= \)
\( \phantom{9.\quad}=e^x(e^x-1)+e^x\cdot e^x=e^x(e^x-1+e^x)=e^x(2e^x-1) \)
\( 10.\quad ((e^x+2)(3-e^x))’=(e^x+2)'(3-e^x)+(e^x+2)(3-e^x)’= \)
\( \phantom{10.\quad}=e^x(3-e^x)+(e^x+2)(-e^x)=e^x(3-e^x-e^x-2) \)
\( \phantom{10.\quad}=e^x(1-2e^x) \)
\( \phantom{6.\quad}=e^x(x+1)+e^x\cdot 1=e^x(x+1+1)=e^x(x+2) \)
\( 7.\quad ((2x-5)e^x)’=(2x-5)’e^x+(2x-5)(e^x)’= \)
\( \phantom{7.\quad}=2e^x+(2x-5)e^x=e^x(2+2x-5)=e^x(2x-3) \)
\( 8.\quad (e^x(3x^2-2x-5))’=(e^x)'(3x^2-2x-5)+e^x(3x^2-2x-5)’= \)
\( \phantom{8.\quad}=e^x(3x^2-2x-5)+e^x(6x-2)=e^x(3x^2-2x-5+6x-2) \)
\( \phantom{8.\quad}=e^x(3x^2+4x-7) \)
\( 9.\quad (e^x(e^x-1))’=(e^x)'(e^x-1)+e^x(e^x-1)’= \)
\( \phantom{9.\quad}=e^x(e^x-1)+e^x\cdot e^x=e^x(e^x-1+e^x)=e^x(2e^x-1) \)
\( 10.\quad ((e^x+2)(3-e^x))’=(e^x+2)'(3-e^x)+(e^x+2)(3-e^x)’= \)
\( \phantom{10.\quad}=e^x(3-e^x)+(e^x+2)(-e^x)=e^x(3-e^x-e^x-2) \)
\( \phantom{10.\quad}=e^x(1-2e^x) \)
\(
11.\quad (x\ln x)’=x’\ln x+x(\ln x)’=
\)
\( \phantom{11.\quad}=\ln x+x\cdot\frac{1}{x}=\ln x+1 \)
\( 12.\quad (x^2\ln x)’=(x^2)’\ln x+x^2(\ln x)’= \)
\( \phantom{12.\quad}=2x\ln x+x^2\cdot\frac{1}{x}=2x\ln x+x \)
\( 13.\quad ((2x-5)\ln x)’=(2x-5)’\ln x+(2x-5)(\ln x)’= \)
\( \phantom{13.\quad}=2\ln x+(2x-5)\cdot\frac{1}{x}=2\ln x+\frac{2x-5}{x} \)
\( 14.\quad (\ln x(x^2+x+1))’=(\ln x)'(x^2+x+1)+\ln x(x^2+x+1)’= \)
\( \phantom{14.\quad}=\frac{1}{x}(x^2+x+1)+\ln x(2x+1) \)
\( \phantom{11.\quad}=\ln x+x\cdot\frac{1}{x}=\ln x+1 \)
\( 12.\quad (x^2\ln x)’=(x^2)’\ln x+x^2(\ln x)’= \)
\( \phantom{12.\quad}=2x\ln x+x^2\cdot\frac{1}{x}=2x\ln x+x \)
\( 13.\quad ((2x-5)\ln x)’=(2x-5)’\ln x+(2x-5)(\ln x)’= \)
\( \phantom{13.\quad}=2\ln x+(2x-5)\cdot\frac{1}{x}=2\ln x+\frac{2x-5}{x} \)
\( 14.\quad (\ln x(x^2+x+1))’=(\ln x)'(x^2+x+1)+\ln x(x^2+x+1)’= \)
\( \phantom{14.\quad}=\frac{1}{x}(x^2+x+1)+\ln x(2x+1) \)
\(
15.\quad (x\sqrt{x})’=x’\sqrt{x}+x(\sqrt{x})’=
\)
\( \phantom{15.\quad}=\sqrt{x}+x\cdot\frac{1}{2\sqrt{x}}=\sqrt{x}+\frac{x}{2\sqrt{x}}=\sqrt{x}+\frac{\sqrt{x}}{2}=\frac{3\sqrt{x}}{2} \)
\( 16.\quad \left(x^2\sqrt[3]{x}\right)’=(x^2)’\sqrt[3]{x}+x^2(\sqrt[3]{x})’= \)
\( \phantom{16.\quad}=2x\sqrt[3]{x}+x^2\cdot\frac{1}{3\sqrt[3]{x^2}}=2x\sqrt[3]{x}+\frac{x\sqrt[3]{x}}{3}=\frac{7x\sqrt[3]{x}}{3} \)
\( 17.\quad (x^2\sin x)’=(x^2)’\sin x+x^2(\sin x)’= \)
\( \phantom{17.\quad}=2x\sin x+x^2\cos x \)
\( 18.\quad (x\cos x)’=x’\cos x+x(\cos x)’= \)
\( \phantom{18.\quad}=\cos x-x\sin x \)
\( 19.\quad (\sin^2 x)’=(\sin x\cdot\sin x)’=(\sin x)’\sin x+\sin x(\sin x)’= \)
\( \phantom{19.\quad}=\cos x\sin x+\sin x\cos x=2\sin x\cos x \)
\( 20.\quad (\cos^2 x)’=(\cos x\cdot\cos x)’=(\cos x)’\cos x+\cos x(\cos x)’= \)
\( \phantom{20.\quad}=-\sin x\cos x-\cos x\sin x=-2\sin x\cos x \)
\( \phantom{15.\quad}=\sqrt{x}+x\cdot\frac{1}{2\sqrt{x}}=\sqrt{x}+\frac{x}{2\sqrt{x}}=\sqrt{x}+\frac{\sqrt{x}}{2}=\frac{3\sqrt{x}}{2} \)
\( 16.\quad \left(x^2\sqrt[3]{x}\right)’=(x^2)’\sqrt[3]{x}+x^2(\sqrt[3]{x})’= \)
\( \phantom{16.\quad}=2x\sqrt[3]{x}+x^2\cdot\frac{1}{3\sqrt[3]{x^2}}=2x\sqrt[3]{x}+\frac{x\sqrt[3]{x}}{3}=\frac{7x\sqrt[3]{x}}{3} \)
\( 17.\quad (x^2\sin x)’=(x^2)’\sin x+x^2(\sin x)’= \)
\( \phantom{17.\quad}=2x\sin x+x^2\cos x \)
\( 18.\quad (x\cos x)’=x’\cos x+x(\cos x)’= \)
\( \phantom{18.\quad}=\cos x-x\sin x \)
\( 19.\quad (\sin^2 x)’=(\sin x\cdot\sin x)’=(\sin x)’\sin x+\sin x(\sin x)’= \)
\( \phantom{19.\quad}=\cos x\sin x+\sin x\cos x=2\sin x\cos x \)
\( 20.\quad (\cos^2 x)’=(\cos x\cdot\cos x)’=(\cos x)’\cos x+\cos x(\cos x)’= \)
\( \phantom{20.\quad}=-\sin x\cos x-\cos x\sin x=-2\sin x\cos x \)
\[
\left(\frac{f}{g}\right)’=\frac{f’\cdot g-f\cdot g’}{g^2}
\]
Exemple:
\( 1.\quad \left(\frac{1}{x}\right)’=\frac{1’\cdot x-1\cdot x’}{x^2}= \frac{0\cdot x-1\cdot1}{x^2}=-\frac{1}{x^2} \)\( 2.\quad \left(\frac{1}{\sqrt{x}}\right)’=\frac{1’\sqrt{x}-1(\sqrt{x})’}{(\sqrt{x})^2}= \frac{0\cdot\sqrt{x}-\frac{1}{2\sqrt{x}}}{x}=-\frac{1}{2x\sqrt{x}} \)
\( 3.\quad \left(\frac{1}{e^x}\right)’=\frac{1’e^x-1(e^x)’}{(e^x)^2}= \frac{0\cdot e^x-e^x}{e^{2x}}=-\frac{1}{e^x} \)
\( 4.\quad \left(\frac{1}{2x-1}\right)’=\frac{1′(2x-1)-1(2x-1)’}{(2x-1)^2}= \frac{0\cdot(2x-1)-2}{(2x-1)^2}=-\frac{2}{(2x-1)^2} \)
\( 5.\quad \left(\frac{\sqrt{x}}{x}\right)’=\frac{(\sqrt{x})’x-\sqrt{x}\cdot x’}{x^2}= \frac{\frac{1}{2\sqrt{x}}\cdot x-\sqrt{x}}{x^2}=-\frac{1}{2x\sqrt{x}} \)
\( 6.\quad \left(\frac{x}{x+2}\right)’=\frac{x'(x+2)-x(x+2)’}{(x+2)^2}= \frac{1\cdot(x+2)-x\cdot1}{(x+2)^2}=\frac{2}{(x+2)^2} \)
\( 7.\quad \left(\frac{x}{x^2+3}\right)’=\frac{x'(x^2+3)-x(x^2+3)’}{(x^2+3)^2}= \frac{x^2+3-2x^2}{(x^2+3)^2}=\frac{3-x^2}{(x^2+3)^2} \)
\( 8.\quad \left(\frac{2x-5}{x+2}\right)’=\frac{(2x-5)'(x+2)-(2x-5)(x+2)’}{(x+2)^2}= \frac{2(x+2)-(2x-5)}{(x+2)^2}=\frac{9}{(x+2)^2} \)
\( 9.\quad \left(\frac{x^2-1}{4x-3}\right)’=\frac{(x^2-1)'(4x-3)-(x^2-1)(4x-3)’}{(4x-3)^2}= \)
\( \phantom{9.\quad}=\frac{2x(4x-3)-4(x^2-1)}{(4x-3)^2} =\frac{8x^2-6x-4x^2+4}{(4x-3)^2} =\frac{4x^2-6x+4}{(4x-3)^2} \)
\(
10.\quad \left(\frac{e^x}{2x+1}\right)’=\frac{(e^x)'(2x+1)-e^x(2x+1)’}{(2x+1)^2}=
\frac{e^x(2x+1)-2e^x}{(2x+1)^2}
=\frac{e^x(2x-1)}{(2x+1)^2}
\)
\( 11.\quad \left(\frac{x^2+x-2}{e^x}\right)’=\frac{(x^2+x-2)’e^x-(x^2+x-2)(e^x)’}{(e^x)^2}= \frac{(2x+1)e^x-(x^2+x-2)e^x}{e^{2x}} =\frac{-x^2+x+3}{e^x} \)
\( 12.\quad \left(\frac{1-e^x}{1+e^x}\right)’=\frac{(1-e^x)'(1+e^x)-(1-e^x)(1+e^x)’}{(1+e^x)^2}= \frac{(-e^x)(1+e^x)-(1-e^x)e^x}{(1+e^x)^2} =\frac{-2e^x}{(1+e^x)^2} \)
\( 13.\quad \left(\frac{\ln x}{x}\right)’=\frac{(\ln x)’x-\ln x\cdot x’}{x^2}= \frac{\frac{1}{x}\cdot x-\ln x}{x^2} =\frac{1-\ln x}{x^2} \)
\( 14.\quad \left(\frac{1-\sin x}{\cos x}\right)’=\frac{(1-\sin x)’\cos x-(1-\sin x)(\cos x)’}{\cos^2 x}= \frac{(-\cos x)\cos x+(1-\sin x)\sin x}{\cos^2 x} =\frac{\sin x-1}{\cos^2 x} \)
\( 15.\quad \left(\frac{\cos x}{2+\sin x}\right)’=\frac{(\cos x)'(2+\sin x)-\cos x(2+\sin x)’}{(2+\sin x)^2}= \frac{(-\sin x)(2+\sin x)-\cos^2 x}{(2+\sin x)^2} =\frac{-2\sin x-1}{(2+\sin x)^2} \)
\( 11.\quad \left(\frac{x^2+x-2}{e^x}\right)’=\frac{(x^2+x-2)’e^x-(x^2+x-2)(e^x)’}{(e^x)^2}= \frac{(2x+1)e^x-(x^2+x-2)e^x}{e^{2x}} =\frac{-x^2+x+3}{e^x} \)
\( 12.\quad \left(\frac{1-e^x}{1+e^x}\right)’=\frac{(1-e^x)'(1+e^x)-(1-e^x)(1+e^x)’}{(1+e^x)^2}= \frac{(-e^x)(1+e^x)-(1-e^x)e^x}{(1+e^x)^2} =\frac{-2e^x}{(1+e^x)^2} \)
\( 13.\quad \left(\frac{\ln x}{x}\right)’=\frac{(\ln x)’x-\ln x\cdot x’}{x^2}= \frac{\frac{1}{x}\cdot x-\ln x}{x^2} =\frac{1-\ln x}{x^2} \)
\( 14.\quad \left(\frac{1-\sin x}{\cos x}\right)’=\frac{(1-\sin x)’\cos x-(1-\sin x)(\cos x)’}{\cos^2 x}= \frac{(-\cos x)\cos x+(1-\sin x)\sin x}{\cos^2 x} =\frac{\sin x-1}{\cos^2 x} \)
\( 15.\quad \left(\frac{\cos x}{2+\sin x}\right)’=\frac{(\cos x)'(2+\sin x)-\cos x(2+\sin x)’}{(2+\sin x)^2}= \frac{(-\sin x)(2+\sin x)-\cos^2 x}{(2+\sin x)^2} =\frac{-2\sin x-1}{(2+\sin x)^2} \)
