Тригонометриялық функциялардың туындысы

$f(x)=2\cos x-{\displaystyle \frac{(\sqrt{\pi }{)}^3}{\sqrt{x}}}+{\displaystyle \frac{\pi }{2}}$
$(\sqrt{\pi }{)}^3$ - коэффициент, ал ${\displaystyle \frac{\pi }{2}}$-тұрақты екенін ескереміз. Сонда,
$f^{\prime }(x)=2(\cos x{)}^{\prime }-(\sqrt{\pi }{)}^3\cdot {\left(x^{-\frac{1}{2}}\right)}^{\prime }+{\left({\displaystyle \frac{\pi }{2}}\right)}^{\prime }=-2\sin x-(\sqrt{\pi }{)}^3\cdot (-{\displaystyle \frac{1}{2}}{x}^{-\frac{3}{2}})+0=$
$=-2\sin x+{\displaystyle \frac{1}{2}}(\sqrt{\pi }{)}^3\cdot {\displaystyle \frac{1}{(\sqrt{x}{)}^3}}=-2\sin x+{\displaystyle \frac{1}{2}}\sqrt{{\displaystyle \frac{{\pi }^3}{x^3}}}$.

$f^{\prime }(x)=\sqrt{3}(\cos x{)}^{\prime }+{(\cos {\displaystyle \frac{\pi }{3}})}^{\prime }+{\displaystyle \frac{3}{\pi }}{\left(x^2\right)}^{\prime }=-\sqrt{3}\sin x+0+{\displaystyle \frac{3}{\pi }}\cdot 2x=$
$=-\sqrt{3}\sin x+{\displaystyle \frac{6x}{\pi }}$
$f^{\prime }\left({\displaystyle \frac{\pi }{3}}\right)=-\sqrt{3}\cdot \sin {\displaystyle \frac{\pi }{3}}+{\displaystyle \frac{6}{\pi }}\cdot {\displaystyle \frac{\pi }{3}}=-\sqrt{3}\cdot {\displaystyle \frac{\sqrt{3}}{2}}+2={\displaystyle \frac{1}{2}}$.

$f^{\prime }(x)=(\sin x{)}^{\prime }\cdot \sqrt{2x}+\sqrt{2}\cdot (\sqrt{x}{)}^{\prime }\cdot \sin x+(2x{)}^{\prime }+(3{)}^{\prime }=$
$=\cos x\cdot \sqrt{2x}+{\displaystyle \frac{\sqrt{2}}{2\sqrt{x}}}\cdot \sin x+2+0=\sqrt{2x}\cos x+{\displaystyle \frac{\sin x}{\sqrt{2x}}}+2$
$f^{\prime }\left({\displaystyle \frac{\pi }{2}}\right)=\sqrt{2\cdot {\displaystyle \frac{\pi }{2}}}\cdot \cos {\displaystyle \frac{\pi }{2}}+{\displaystyle \frac{\sin \frac{\pi }{2}}{\sqrt{2\cdot \frac{\pi }{2}}}}+2=0+{\displaystyle \frac{1}{\sqrt{\pi }}}+2={\displaystyle \frac{\sqrt{\pi }}{\pi }}+2$

$f^{\prime }(x)={\displaystyle \frac{(\mathrm{tg}x{)}^{\prime }\cdot x-\mathrm{tg}x\cdot x^{\prime }}{x^2}}={\displaystyle \frac{\frac{x}{{\cos }^{2}x}-\mathrm{tg}x}{x^2}}={\displaystyle \frac{\frac{x}{{\cos }^{2}x}-\frac{\sin x}{\cos x}}{x^2}}={\displaystyle \frac{\frac{x-\sin x\cos x}{{\cos }^{2}x}}{x^2}}=$
$={\displaystyle \frac{x-\sin x\cos x}{x^2\cdot {\cos }^{2}x}}$

$f(x)={\displaystyle \frac{\mathrm{tg}x+1}{\mathrm{tg}x}}={\displaystyle \frac{\mathrm{tg}x}{\mathrm{tg}x}}+{\displaystyle \frac{1}{\mathrm{tg}x}}=1+\mathrm{ctg}x$
$f^{\prime }(x)=(1+\mathrm{ctg}x{)}^{\prime }=-{\displaystyle \frac{1}{{\sin }^{2}x}}$

$f(x)=3\sin ({\displaystyle \frac{\pi }{2}}+2x)-{\displaystyle \frac{x+{\pi }^2}{x}}=3\cos 2x-({\displaystyle \frac{x}{x}}+{\displaystyle \frac{{\pi }^2}{x}})=3\cos 2x-1-{\displaystyle \frac{{\pi }^2}{x}}$
$f^{\prime }(x)=3(\cos 2x{)}^{\prime }-(1{)}^{\prime }-{\pi }^2{\left({\displaystyle \frac{1}{x}}\right)}^{\prime }=3(-\sin 2x)\cdot (2x{)}^{\prime }-0-{\pi }^2(-{\displaystyle \frac{1}{x^2}})=$
$=-6\sin 2x+{\left({\displaystyle \frac{\pi }{x}}\right)}^2$
$f^{\prime }\left({\displaystyle \frac{\pi }{12}}\right)=-6\sin {\displaystyle \frac{\pi }{6}}+{12}^2=-3+144=141$

$f(x)={\sin }^{4}x-{\cos }^{4}x=({\sin }^{2}x+{\cos }^{2}x)({\sin }^{2}x-{\cos }^{2}x)=-\cos 2x$
$f^{\prime }(x)=-(\cos 2x{)}^{\prime }=-(-\sin 2x)\cdot (2x{)}^{\prime }=2\sin 2x$
$f^{\prime }\left({\displaystyle \frac{\pi }{12}}\right)=2\sin {\displaystyle \frac{\pi }{6}}=1$

$f(x)={\displaystyle \frac{(2\sin x\cdot \cos x{)}^2}{4}}={\displaystyle \frac{{\sin }^{2}2x}{4}}={\displaystyle \frac{1-\cos 4x}{4\cdot 2}}={\displaystyle \frac{1}{8}}-{\displaystyle \frac{1}{8}}\cos 4x$
$f^{\prime }(x)={({\displaystyle \frac{1}{8}}-{\displaystyle \frac{1}{8}}\cos 4x)}^{\prime }=-{\displaystyle \frac{1}{8}}(\cos 4x{)}^{\prime }=-{\displaystyle \frac{1}{8}}\cdot (-\sin 4x)\cdot (4x{)}^{\prime }={\displaystyle \frac{1}{2}}\sin 4x$

$f^{\prime }(x)=(\sin (\cos x){)}^{\prime }=\cos (\cos x)\cdot (\cos x{)}^{\prime }=-\sin x\cdot \cos (\cos x)$

$h^{\prime }(x)={\left(\mathrm{tg}(2x-3x^2)\right)}^{\prime }={\displaystyle \frac{1}{{\cos }^2(2x-3x^2)}}\cdot {(2x-3x^2)}^{\prime }={\displaystyle \frac{2-6x}{{\cos }^2(2x-3x^2)}}$

$f^{\prime }(x)={\left(\mathrm{ctg}\left({\displaystyle \frac{x+1}{2}}\right)\right)}^{\prime }=-{\displaystyle \frac{1}{{\sin }^2\left(\frac{x+1}{2}\right)}}\cdot {\left({\displaystyle \frac{x+1}{2}}\right)}^{\prime }=-{\displaystyle \frac{1}{{\sin }^2\left(\frac{x+1}{2}\right)}}\cdot {({\displaystyle \frac{1}{2}}x+{\displaystyle \frac{1}{2}})}^{\prime }=$
$=-{\displaystyle \frac{1}{{\sin }^2\left(\frac{x+1}{2}\right)}}\cdot {\displaystyle \frac{1}{2}}=-{\displaystyle \frac{1}{2{\sin }^2\left(\frac{x+1}{2}\right)}}$

$f^{\prime }(x)=2\cos \sqrt[3]{x}\cdot (\cos \sqrt[3]{x}{)}^{\prime }=2\cos \sqrt[3]{x}\cdot (-\sin \sqrt[3]{x})\cdot (\sqrt[3]{x}{)}^{\prime }=$
$=-2\cos \sqrt[3]{x}\cdot \sin \sqrt[3]{x}\cdot {\displaystyle \frac{1}{3}}{x}^{-\frac{2}{3}}=-{\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{1}{\sqrt[3]{x^2}}}\cdot \sin (2\sqrt[3]{x})=-{\displaystyle \frac{\sqrt[3]{x}\cdot \sin (2\sqrt[3]{x})}{3x}}$

$f(x)={\cos }^3{\displaystyle \frac{x}{3}}+\mathrm{ctg}({\displaystyle \frac{\pi }{4}}-x)+{\sin }^2{\displaystyle \frac{\pi }{13}}={(\cos {\displaystyle \frac{x}{3}})}^3+\mathrm{ctg}({\displaystyle \frac{\pi }{4}}-x)+{\sin }^2{\displaystyle \frac{\pi }{13}}$
$f^{\prime }(x)={\left({(\cos {\displaystyle \frac{x}{3}})}^3\right)}^{\prime }+{\left(\mathrm{ctg}({\displaystyle \frac{\pi }{4}}-x)\right)}^{\prime }+{({\sin }^2{\displaystyle \frac{\pi }{13}})}^{\prime }=3{\cos }^2{\displaystyle \frac{x}{3}}\cdot {(\cos {\displaystyle \frac{x}{3}})}^{\prime }-$
$-{\displaystyle \frac{1}{{\sin }^2(\frac{\pi }{4}-x)}}\cdot {({\displaystyle \frac{\pi }{4}}-x)}^{\prime }=3{\cos }^2{\displaystyle \frac{x}{3}}\cdot (-\sin {\displaystyle \frac{x}{3}})\cdot {\left({\displaystyle \frac{x}{3}}\right)}^{\prime }-{\displaystyle \frac{1}{{\sin }^2(\frac{\pi }{4}-x)}}\cdot (-1)=$
$=-{\cos }^2{\displaystyle \frac{x}{3}}\cdot \sin {\displaystyle \frac{x}{3}}+{\displaystyle \frac{1}{{\sin }^2(x-\frac{\pi }{4})}}={\displaystyle \frac{1}{{\sin }^2(x-\frac{\pi }{4})}}-{\cos }^2{\displaystyle \frac{x}{3}}\cdot \sin {\displaystyle \frac{x}{3}}$
$f^{\prime }\left({\displaystyle \frac{3\pi }{4}}\right)={\displaystyle \frac{1}{{\sin }^2\frac{\pi }{2}}}-{\cos }^2{\displaystyle \frac{\pi }{4}}\sin {\displaystyle \frac{\pi }{4}}=1-{\displaystyle \frac{\sqrt{2}}{4}}$