Derivace

$y^{\prime} \; = \; \Large ( \normalsize 3x^7-{\Large\frac{2x^2-1}{4})^{\prime}} \; = \; (3x^7)^{\prime}-{\Large (\frac{2x^2-1}{4})^{\prime}} \; =$

$= \; 3(x^7)^{\prime}-{\Large ( \frac{1}{4}}\cdot (2x^2-1) {\Large )^{\prime}} \; = \; 3\cdot 7x^6-{\Large\frac{1}{4}}\cdot (2x^2-1)^{\prime} \; =$

$= 21 x^6-{\Large\frac{1}{4}} (2(x^2)^{\prime}-(1)^{\prime}) \; = \; 21 x^6-{\Large\frac{1}{4}} (2\cdot 2x-0) \; = 21x^6-x$

$y^{\prime}= (e^x-x+2)^{\prime}=$$\; (e^x)^{\prime}-(x)^{\prime}+(2)^{\prime} =$$\; e^x-1+0 = e^x-1$

$y^{\prime}= (2e^x+{\Large\frac{x^3}{3}}-2x^2-3\pi)^{\prime}=$$\; (2e^x)^{\prime}+({\Large\frac{x^3}{3}})^{\prime}-(2x^2)^{\prime}-(3\pi)^{\prime} =$

$= 2(e^x)^{\prime}+{\Large\frac{(x^3)^{\prime}}{3}}-2(x^2)^{\prime}-(3\pi)^{\prime} =$$\; 2e^x+{\Large\frac{3x^2}{3}}-2\cdot 2x-0 =$$\; 2e^x+x^2-4x$

$y^{\prime} = {\Large (}\pi x^4-{\Large\frac{x^2+2x}{2})^{\prime}}=$$\; (\pi x^4)^{\prime}-{\Large ( \frac{x^2+2x}{2} )^{\prime}} =$$\; \pi (x^4)^{\prime}-{\Large\frac{(x^2+2x)^{\prime}}{2}} =$

$= \pi (x^4)^{\prime}-{\Large\frac{(x^2)^{\prime}+(2x)^{\prime}}{2}} =$$\; \pi\cdot 4 x^3-{\Large\frac{2x+2}{2}} =$$\; 4\pi x^3-x-1$

$y^{\prime} \; = \; (3xe^x+x^2-1)^{\prime} \; = \; 3(xe^x)^{\prime}+(x^2)^{\prime}-(1)^{\prime} \; = \; 3[x(e^x)^{\prime}+(x)^{\prime}e^x] \; +$

$+ \; 2x-0 \; = \; 3[xe^x+1\cdot e^x] + 2x \; = \; 3e^x(x+1)+2x$

$y^{\prime}= [(x^2-1)e^x]^{\prime}=$$\; (x^2-1)^{\prime}e^x+(x^2-1)(e^x)^{\prime} =$$\; [(x^2)^{\prime}-(1)^{\prime}]e^x \; +$

$+\; (x^2-1)e^x =$$\; [2x-0]e^x+(x^2-1)e^x =$$\; 2xe^x+(x^2-1)e^x =$

$= (x^2+2x-1)e^x$

$y^{\prime}= [(-x^2+2x)e^x]^{\prime}=$$\; (-x^2+2x)^{\prime}e^x+(-x^2+2x)(e^x)^{\prime} =$

$= [-(x^2)^{\prime}+2(x)^{\prime}]e^x + (-x^2+2x)e^x =$$\; (-2x+2)e^x+(-x^2+2x)e^x =$

$= (-x^2+2)e^x$

$y^{\prime}= (e^x\cdot e^x)^{\prime}=$$\; (e^x)^{\prime}e^x+e^x(e^x)^{\prime} =$$\; e^x\cdot e^x + e^x\cdot e^x =$$\; 2e^x\cdot e^x =$$\; 2e^{2x}$

$y^{\prime} \; = \; (\sin{x}\cos{x})^{\prime} \; = \; (\sin{x})^{\prime}\cos{x}+\sin{x}(\cos{x})^{\prime} \; = \; \cos{x}\cos{x} \; +$

$+ \; \sin{x}(-\sin{x}) \; = \; \cos^2{x} - \sin^2{x} \; = \; \cos{2x}$

$y^{\prime} \; = \; [e^x(\sin{x}+\cos{x})]^{\prime} \; = \; (e^x)^{\prime}(\sin{x}+\cos{x}) + e^x(\sin{x}+\cos{x})^{\prime} \; =$

$= \; e^x(\sin{x}+\cos{x}) + e^x[(\sin{x})^{\prime}+(\cos{x})^{\prime}] \; =$

$= \; e^x(\sin{x}+\cos{x}) + e^x(\cos{x}-\sin{x}) \; = \; 2e^x\cos{x}$

$y^{\prime}= (3\cos{x}-4\sin{x}+3^x)^{\prime}=$$\; 3(\cos{x})^{\prime}-4(\sin{x})^{\prime}+(3^x)^{\prime} =$

$= 3(-\sin{x}) -4\cos{x}+3^x\ln{3} =$$\; -3\sin{x} -4\cos{x}+3^x\ln{3}$

$y^{\prime}= (\sin{x} \cdot \sin{x})^{\prime}=$$\; (\sin{x})^{\prime}\sin{x}+\sin{x}(\sin{x})^{\prime} =$$\; 2\sin{x}(\sin{x})^{\prime} =$

$= 2\sin{x}\cos{x} =$$\; \sin{2x}$

$y^{\prime}= (\cos{x} \cdot \cos{x})^{\prime}=$$\; (\cos{x})^{\prime}\cos{x}+\cos{x}(\cos{x})^{\prime} =$$\; 2(\cos{x})^{\prime}\cos{x} =$

$= 2(-\sin{x})\cos{x} =$$\; -2\sin{x}\cos{x} =$$\; -\sin{2x}$

$y^{\prime}= (2^x \cdot 3^x)^{\prime}=$$\; (2^x)^{\prime}3^x+2^x(3^x)^{\prime} =$$\; 2^x(\ln{2})3^x+2^x3^x(\ln{3}) =$

$= 2^x3^x(\ln{2}+\ln{3}) =$$\; 2^x3^x\ln(2\cdot 3) =$$\; 6^x\ln{6}$

$y^{\prime}= (2^x \cdot 3^x)^{\prime}=$$\; (6^x)^{\prime} =$$\; 6^x\ln{6}$

$u(x) = \sqrt{x} \;$ a $\; v(x) = x^3+2$.

Vnější funkce $u(x) = \sqrt{x} = x^{\frac{1}{2}}$ je elementární funkce. Je to mocninná funkce s exponentem jedna polovina.

$u(x) = x^3 \;$ a $\; v(x) = \sin{x}$.

Vnější funkce $u(x) = x^3$ je elementární funkce. Je to mocninná funkce s exponentem tři.

$u(x) = \Large\frac{2}{x} \;$ a $\; v(x) = x^2+3\cos{x}$.

Vnější funkce $u(x) = \Large\frac{2}{x}\normalsize = 2x^{-1}$ je elementární funkce násobená konstantou. Je to mocninná funkce s exponentem mínus jedna.

$u(x) = -\pi\cos{x} \;$ a $\; v(x) = 3+e^x$.

Vnější funkce $u(x) = -\pi\cos{x}$ je elementární funkce násobená konstantou. Je to funkce kosinus.

Vnější funkce je $u(x) = \sin{x}$. Vnitřní funkce je $v(x) = x^2+\Large\frac{\pi}{2}$.

$y^{\prime} \; = \; [\sin(x^2+\Large\frac{\pi}{2}\normalsize)]^{\prime} \; = \; \sin^{\prime}(x^2+\Large\frac{\pi}{2}\normalsize)\cdot (x^2+\Large\frac{\pi}{2}\normalsize)^{\prime} \; = \; \cos(x^2+\Large\frac{\pi}{2}\normalsize)\cdot 2x \; =$

$= \; 2x\cos(x^2+\Large\frac{\pi}{2}\normalsize) \; = \; -2x\sin(x^2)$

Vnější funkce je $u(x) = x^5$. Vnitřní funkce je $v(x) = x^3-x$.

$y^{\prime} \; = \; [(x^3-x)^5]^{\prime} \; = \; 5(x^3-x)^4 \cdot (x^3-x)^{\prime} \; = \; 5(x^3-x)^4 (3x^2-1)$

Vnější funkce je $u(x) = 3^x$. Vnitřní funkce je $v(x) = \cos{x}$.

$y^{\prime} \; = \; [3^{\cos{x}}]^{\prime} \; = \; 3^{\cos{x}}(\ln{3})\cdot (\cos{x})^{\prime} \; = \; 3^{\cos{x}}(\ln{3}) \cdot (-\sin{x}) \; =$

$= \; -(\ln{3})\cdot \sin{x} \cdot 3^{\cos{x}}$

Vnější funkce je $u(x) = e^x$. Vnitřní funkce je $v(x) = 3x$.

$y^{\prime}= (e^{3x})^{\prime}=$$\; e^{3x}\cdot (3x)^{\prime} =$$\; e^{3x}\cdot 3 =$$\; 3e^{3x}$

Vnější funkce je $u(x) = \sin{x}$. Vnitřní funkce je $v(x) = e^x$.

$y^{\prime}= (\sin{e^x})^{\prime}=$$\; \sin^{\prime}(e^x)\cdot (e^x)^{\prime} =$$\; (\cos{e^x})\cdot e^x =$$\; e^x\cos{e^x}$

Vnější funkce je $u(x) = x^7$. Vnitřní funkce je $v(x) = x^2+2x-1$.

$y^{\prime}= [(x^2+2x-1)^7]^{\prime}=$$\; 7(x^2+2x-1)^6 \cdot (x^2+2x-1)^{\prime} =$

$= 7(x^2+2x-1)^6 (2x+2) =$$\; 14(x^2+2x-1)^6 (x+1)$

Vnější funkce je $u(x) = x^2$. Vnitřní funkce je $v(x) = \sin(x^3-4x)$.

$y^{\prime}= [\sin^2(x^3-4x)]^{\prime}=$$\; 2\sin(x^3-4x) \cdot \sin^{\prime}(x^3-4x) = (*)$

$\sin^{\prime}(...)$ je opět složená funkce. Vnější funkce je $w_1(x) = \sin{x}$. Vnitřní funkce je $w_2(x) = x^3-4x$.

$(*) = 2\sin(x^3-4x) \cdot \cos(x^3-4x) \cdot (x^3-4x)^{\prime} =$

$= \sin [2(x^3-4x)] \cdot (3x^2-4) =$$\; (3x^2-4)\sin [2(x^3-4x)]$

$y^{\prime} \; = \; \Large[\frac{-2}{x^3-x}]^{\prime}\normalsize \; = \; {-2}\Large[\frac{1}{x^3-x}]^{\prime}\normalsize = \; {-2}\Large\frac{-(x^3-x)^{\prime}}{(x^3-x)^2}\normalsize \; = \; \Large\frac{2(3x^2-1)}{(x^3-x)^2}$

$y^{\prime} \; = \; \Large[\frac{-3x+1}{x^2+2}]^{\prime}\normalsize \; = \; \Large\frac{(-3x+1)^{\prime}(x^2+2)-(-3x+1)(x^2+2)^{\prime}}{(x^2+2)^2}\normalsize =$

$= \; \Large\frac{-3(x^2+2)-(-3x+1)2x}{(x^2+2)^2}\normalsize \; = \; \Large\frac{-3x^2-6-2x(-3x+1)}{(x^2+2)^2}\normalsize \; =$

$= \; \Large\frac{-3x^2-6+6x^2-2x}{(x^2+2)^2}\normalsize \; = \; \Large\frac{3x^2-2x-6}{(x^2+2)^2}$

$y^{\prime}= [x + \Large\frac{1}{2x^2}\normalsize]^{\prime}=$$\; (x)^{\prime}+\Large\frac{1}{2}\normalsize (x^{-2})^{\prime} =$$\; 1+\Large\frac{1}{2}\normalsize (-2)x^{-3} =$$\; 1-\Large\frac{1}{x^3}$

$y^{\prime}= \Large[\frac{3x+1}{x^2-2}]^{\prime}\normalsize =$$\; \Large\frac{(3x+1)^{\prime}(x^2-2)-(3x+1)(x^2-2)^{\prime}}{(x^2-2)^2}\normalsize =$

$= \Large\frac{3(x^2-2)-(3x+1)2x}{(x^2-2)^2}\normalsize =$$\; \Large\frac{3x^2-6-2x(3x+1)}{(x^2-2)^2}\normalsize \; =$

$= \Large\frac{3x^2-6-6x^2-2x}{(x^2-2)^2}\normalsize =$$\; \Large\frac{-3x^2-2x-6}{(x^2-2)^2}$

$y^{\prime}= ({\rm tg}\:x - {\rm cotg}\:x)^{\prime} =$$\; ({\rm tg}\:x)^{\prime} - ({\rm cotg}\:x)^{\prime} =$$\; \Large\frac{1}{\cos^2{x}}\normalsize - (-\Large\frac{1}{\sin^2{x}})\normalsize =$

$= \Large\frac{1}{\cos^2{x}}\normalsize + \Large\frac{1}{\sin^2{x}}\normalsize =$$\; \Large\frac{\sin^2{x}+\cos^2{x}}{\sin^2{x}\cdot\cos^2{x}}\normalsize =$$\; \Large\frac{1}{\sin^2{x}\cdot\cos^2{x}}\normalsize=$$\; \Large\frac{4}{(2\sin{x}\cdot\cos{x})^2}=$

$= \Large\frac{4}{\sin^2{2x}}$

$y^{\prime}= \Large[\frac{{\rm tg}\:x}{x}]^{\prime}\normalsize =$$\; \Large\frac{({\rm tg}\:x)^{\prime}x-({\rm tg}\:x)(x)^{\prime}}{x^2}\normalsize =$$\; \Large\frac{(\cos^{-2}{x})\cdot x-({\rm tg}\:x)\cdot 1}{x^2}\normalsize =$

$= \Large\frac{1}{x\cos^2{x}}\normalsize-\Large\frac{{\rm tg}\:x}{x^2}$

$y^{\prime} \; = \; [\ln(1+e^x)]^{\prime} \; = \; [\ln^{\prime}(1+e^x)]\cdot (1+e^x)^{\prime} \; = \; \Large\frac{1}{1+e^x}\normalsize\cdot e^x \; = \; \Large\frac{e^x}{1+e^x}$

$y^{\prime} \; = \; [\log_2(2^x+2^{-x})]^{\prime} \; = \; [\log_2^{\prime}(2^x+2^{-x})]\cdot (2^x+2^{-x})^{\prime} \; =$

$= \; \Large\frac{1}{(2^x+2^{-x})\ln{2}}\normalsize\cdot ((2^x)^{\prime}+(2^{-x})^{\prime}) \; = \; \Large\frac{(2^x)^{\prime}+(2^{-x})^{\prime}}{(2^x+2^{-x})\ln{2}}\normalsize \; =$

$= \; \Large\frac{2^x\ln{2} \; + \; 2^{-x}\ln{2}\cdot(-x)^{\prime}}{(2^x+2^{-x})\ln{2}}\normalsize\; = \; \Large\frac{(2^x \; + \; 2^{-x}\cdot(-1))\ln{2}}{(2^x+2^{-x})\ln{2}}\normalsize\; = \; \Large\frac{2^x - 2^{-x}}{2^x+2^{-x}}$

$y^{\prime}= (\ln^2{x})^{\prime}=$$\; 2(\ln{x})\cdot(\ln{x})^{\prime} =$$\; 2(\ln{x})\cdot\Large\frac{1}{x} \normalsize =$$\; \Large\frac{2\ln{x}}{x}$

$y^{\prime}= [\log_2(1+2^x)]^{\prime}=$$\; \Large\frac{1}{(1+2^x)\ln{2}} \normalsize \cdot (1+2^x)^{\prime}=$$\;\Large\frac{1}{(1+2^x)\ln{2}} \normalsize \cdot 2^x\ln{2} =$

$= \Large\frac{2^x}{1+2^x}$

$y^{\prime} = [\log_3(3^x+3^{-x})]^{\prime} =$$\; [\log_3^{\prime}(3^x+3^{-x})]\cdot (3^x+3^{-x})^{\prime} =$

$= \Large\frac{1}{(3^x+3^{-x})\ln{3}}\normalsize\cdot ((3^x)^{\prime}+(3^{-x})^{\prime}) =$$\; \Large\frac{(3^x)^{\prime}+(3^{-x})^{\prime}}{(3^x+3^{-x})\ln{3}}\normalsize =$

$= \Large\frac{3^x\ln{3} \; + \; 3^{-x}\ln{3}\cdot(-x)^{\prime}}{(3^x+3^{-x})\ln{3}}\normalsize =$$\; \Large\frac{(3^x \; + \; 3^{-x}\cdot(-1))\ln{3}}{(3^x+3^{-x})\ln{3}}\normalsize =$$\; \Large\frac{3^x - 3^{-x}}{3^x+3^{-x}}$

$\Large y^{\prime} \; = \; [\sqrt[3]{x^2+2}]^{\prime} \; = \; \Large[(x^2+2)^\frac{1}{3}]^{\prime} \; =$

$\Large = \; \frac{1}{3}(x^2+2)^{-\frac{2}{3}}\cdot (x^2+2)^{\prime} \; =\; \frac{1}{3}(x^2+2)^{-\frac{2}{3}}\cdot 2x \; =$

$\Large = \; \huge\frac{2x}{3\sqrt[3]{(x^2+2)^2}}$

$\Large y^{\prime} = [4\sqrt[4]{2+\cos{x}}]^{\prime} =$$\Large \; 4[(2+\cos{x})^\frac{1}{4}]^{\prime} =$

$\Large = 4\cdot\frac{1}{4}(2+\cos{x})^{-\frac{3}{4}}\cdot (2+\cos{x})^{\prime} =$

$\Large = (2+\cos{x})^{-\frac{3}{4}}\cdot (-\sin{x}) =$$\Large \; -\huge\frac{\sin{x}}{\sqrt[4]{(2+\cos{x})^3}}$

$\Large y^{\prime} = [\sin(3+\sqrt{2x})]^{\prime} =$$\Large \; \sin^{\prime}(3+\sqrt{2x})\cdot [3+\sqrt{2x}]^{\prime} =$

$\Large = \cos(3+\sqrt{2x})\cdot [(2x)^\frac{1}{2}]^{\prime} =$

$\Large = \cos(3+\sqrt{2x})\cdot \frac{1}{2}(2x)^{-\frac{1}{2}}\cdot (2x)^{\prime} =$

$\Large = \cos(3+\sqrt{2x})\cdot \frac{1}{2}(2x)^{-\frac{1}{2}}\cdot 2 =$$\Large \; \huge\frac{\cos(3+\sqrt{2x})}{\sqrt{2x}}$