Тригонометриялық функциялардың туындысы
2.6 Келесі функциялардың туындыларын есептеңіз.
1) $f(x)=2 \cos x-\dfrac{(\sqrt{\pi})^3}{\sqrt{x}}+\dfrac{\pi}{2}$.
Шешуі
$f(x)=2 \cos x-\dfrac{(\sqrt{\pi})^3}{\sqrt{x}}+\dfrac{\pi}{2}$
$(\sqrt{\pi})^3$ – коэффициент, ал $\dfrac{\pi}{2}$-тұрақты екенін ескереміз. Сонда,
$f^{\prime}(x)=2(\cos x)^{\prime}-(\sqrt{\pi})^3 \cdot\left(x^{-\frac{1}{2}}\right)^{\prime}+\left(\dfrac{\pi}{2}\right)^{\prime}=-2 \sin x-(\sqrt{\pi})^3 \cdot\left(-\dfrac{1}{2} x^{-\frac{3}{2}}\right)+0=$
$=-2 \sin x+\dfrac{1}{2}(\sqrt{\pi})^3 \cdot \dfrac{1}{(\sqrt{x})^3}=-2 \sin x+\dfrac{1}{2} \sqrt{\dfrac{\pi^3}{x^3}}$.
2) $f(x)=\sqrt{3} \cos x+\cos \dfrac{\pi}{3}+\dfrac{3}{\pi} x^2, \quad x=\dfrac{\pi}{3}$.
Шешуі
$f^{\prime}(x)=\sqrt{3}(\cos x)^{\prime}+\left(\cos \dfrac{\pi}{3}\right)^{\prime}+\dfrac{3}{\pi}\left(x^2\right)^{\prime}=-\sqrt{3} \sin x+0+\dfrac{3}{\pi} \cdot 2 x=$
$=-\sqrt{3} \sin x+\dfrac{6 x}{\pi}$
$f^{\prime}\left(\dfrac{\pi}{3}\right)=-\sqrt{3} \cdot \sin \dfrac{\pi}{3}+\dfrac{6}{\pi} \cdot \dfrac{\pi}{3}=-\sqrt{3} \cdot \dfrac{\sqrt{3}}{2}+2=\dfrac{1}{2}$.
3) $f(x)=\sin x \cdot \sqrt{2 x}+2 x+3, \quad x=\dfrac{\pi}{2}$
Шешуі
$f^{\prime}(x)=(\sin x)^{\prime} \cdot \sqrt{2 x}+\sqrt{2} \cdot(\sqrt{x})^{\prime} \cdot \sin x+(2 x)^{\prime}+(3)^{\prime}=$
$=\cos x \cdot \sqrt{2 x}+\dfrac{\sqrt{2}}{2 \sqrt{x}} \cdot \sin x+2+0=\sqrt{2 x} \cos x+\dfrac{\sin x}{\sqrt{2 x}}+2$
$f^{\prime}\left(\dfrac{\pi}{2}\right)=\sqrt{2 \cdot \dfrac{\pi}{2}} \cdot \cos \dfrac{\pi}{2}+\dfrac{\sin \frac{\pi}{2}}{\sqrt{2 \cdot \frac{\pi}{2}}}+2=0+\dfrac{1}{\sqrt{\pi}}+2=\dfrac{\sqrt{\pi}}{\pi}+2$
4) $f(x)=\dfrac{\operatorname{tg} x}{x}$.
Шешуі
$f^{\prime}(x)=\dfrac{(\operatorname{tg} x)^{\prime} \cdot x-\operatorname{tg} x \cdot x^{\prime}}{x^2}=\dfrac{\frac{x}{\cos ^2 x}-\operatorname{tg} x}{x^2}=\dfrac{\frac{x}{\cos ^2 x}-\frac{\sin x}{\cos x}}{x^2}=\dfrac{\frac{x-\sin x \cos x}{\cos ^2 x}}{x^2}=$
$=\dfrac{x-\sin x \cos x}{x^2 \cdot \cos ^2 x}$
5) $f(x)=\dfrac{\operatorname{tg} x+1}{\operatorname{tg} x}$.
Шешуі
$f(x)=\dfrac{\operatorname{tg} x+1}{\operatorname{tg} x}=\dfrac{\operatorname{tg} x}{\operatorname{tg} x}+\dfrac{1}{\operatorname{tg} x}=1+\operatorname{ctg} x$
$f^{\prime}(x)=(1+\operatorname{ctg} x)^{\prime}=-\dfrac{1}{\sin ^2 x}$
2.7 Күрделі функцияның туындысын табыңыз.
1) $f(x)=3 \sin \left(2 x+\dfrac{\pi}{2}\right)-\dfrac{x+\pi^2}{x}, \quad f^{\prime}\left(\dfrac{\pi}{12}\right)-?$
Шешуі
$f(x)=3 \sin \left(\dfrac{\pi}{2}+2 x\right)-\dfrac{x+\pi^2}{x}=3 \cos 2 x-\left(\dfrac{x}{x}+\dfrac{\pi^2}{x}\right)=3 \cos 2 x-1-\dfrac{\pi^2}{x}$
$f^{\prime}(x)=3(\cos 2 x)^{\prime}-(1)^{\prime}-\pi^2\left(\dfrac{1}{x}\right)^{\prime}=3(-\sin 2 x) \cdot(2 x)^{\prime}-0-\pi^2\left(-\dfrac{1}{x^2}\right)=$
$=-6 \sin 2 x+\left(\dfrac{\pi}{x}\right)^2$
$f^{\prime}\left(\dfrac{\pi}{12}\right)=-6 \sin \dfrac{\pi}{6}+12^2=-3+144=141$
2) $f(x)=\sin ^4 x-\cos ^4 x, \quad f^{\prime}\left(\dfrac{\pi}{12}\right)-?$
Шешуі
$f(x)=\sin ^4 x-\cos ^4 x=\left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^2 x-\cos ^2 x\right)=-\cos 2 x$
$f^{\prime}(x)=-(\cos 2 x)^{\prime}=-(-\sin 2 x) \cdot(2 x)^{\prime}=2 \sin 2 x$
$f^{\prime}\left(\dfrac{\pi}{12}\right)=2 \sin \dfrac{\pi}{6}=1$
3) $f(x)=\sin ^2 x \cdot \cos ^2 x$.
Шешуі
$f(x)=\dfrac{(2 \sin x \cdot \cos x)^2}{4}=\dfrac{\sin ^2 2 x}{4}=\dfrac{1-\cos 4 x}{4 \cdot 2}=\dfrac{1}{8}-\dfrac{1}{8} \cos 4 x$
$f^{\prime}(x)=\left(\dfrac{1}{8}-\dfrac{1}{8} \cos 4 x\right)^{\prime}=-\dfrac{1}{8}(\cos 4 x)^{\prime}=-\dfrac{1}{8} \cdot(-\sin 4 x) \cdot(4 x)^{\prime}=\dfrac{1}{2} \sin 4 x$
4) $f(x)=\sin (\cos x)$.
Шешуі
$f^{\prime}(x)=(\sin (\cos x))^{\prime}=\cos (\cos x) \cdot(\cos x)^{\prime}=-\sin x \cdot \cos (\cos x)$
5) $h(x)=f(g(x))$, егер $f(x)=\operatorname{tg} x, \quad g(x)=2 x-3 x^2$.
Шешуі
$h^{\prime}(x)=\left(\operatorname{tg}\left(2 x-3 x^2\right)\right)^{\prime}=\dfrac{1}{\cos ^2\left(2 x-3 x^2\right)} \cdot\left(2 x-3 x^2\right)^{\prime}=\dfrac{2-6 x}{\cos ^2\left(2 x-3 x^2\right)}$
6) $f(x)=\operatorname{ctg}\left(\dfrac{x+1}{2}\right)$.
Шешуі
$f^{\prime}(x)=\left(\operatorname{ctg}\left(\dfrac{x+1}{2}\right)\right)^{\prime}=-\dfrac{1}{\sin ^2\left(\frac{x+1}{2}\right)} \cdot\left(\dfrac{x+1}{2}\right)^{\prime}=-\dfrac{1}{\sin ^2\left(\frac{x+1}{2}\right)} \cdot\left(\dfrac{1}{2} x+\dfrac{1}{2}\right)^{\prime}=$
$=-\dfrac{1}{\sin ^2\left(\frac{x+1}{2}\right)} \cdot \dfrac{1}{2}=-\dfrac{1}{2 \sin ^2\left(\frac{x+1}{2}\right)}$
7) $f(x)=\cos ^2(\sqrt[3]{x})$.
Шешуі
$f^{\prime}(x)=2 \cos \sqrt[3]{x} \cdot(\cos \sqrt[3]{x})^{\prime}=2 \cos \sqrt[3]{x} \cdot(-\sin \sqrt[3]{x}) \cdot(\sqrt[3]{x})^{\prime}=$
$=-2 \cos \sqrt[3]{x} \cdot \sin \sqrt[3]{x} \cdot \dfrac{1}{3} x^{-\frac{2}{3}}=-\dfrac{1}{3} \cdot \dfrac{1}{\sqrt[3]{x^2}} \cdot \sin (2 \sqrt[3]{x})=-\dfrac{\sqrt[3]{x} \cdot \sin (2 \sqrt[3]{x})}{3 x}$
8) $f(x)=\cos ^3 \dfrac{x}{3}+\operatorname{ctg}\left(\dfrac{\pi}{4}-x\right)+\sin ^2 \dfrac{\pi}{13}, \quad f^{\prime}\left(\dfrac{3 \pi}{4}\right)-?$
Шешуі
$f(x)=\cos ^3 \dfrac{x}{3}+\operatorname{ctg}\left(\dfrac{\pi}{4}-x\right)+\sin ^2 \dfrac{\pi}{13}=\left(\cos \dfrac{x}{3}\right)^3+\operatorname{ctg}\left(\dfrac{\pi}{4}-x\right)+\sin ^2 \dfrac{\pi}{13}$
$f^{\prime}(x)=\left(\left(\cos \dfrac{x}{3}\right)^3\right)^{\prime}+\left(\operatorname{ctg}\left(\dfrac{\pi}{4}-x\right)\right)^{\prime}+\left(\sin ^2 \dfrac{\pi}{13}\right)^{\prime}=3 \cos ^2 \dfrac{x}{3} \cdot\left(\cos \dfrac{x}{3}\right)^{\prime}-$
$-\dfrac{1}{\sin ^2\left(\frac{\pi}{4}-x\right)} \cdot\left(\dfrac{\pi}{4}-x\right)^{\prime}=3 \cos ^2 \dfrac{x}{3} \cdot\left(-\sin \dfrac{x}{3}\right) \cdot\left(\dfrac{x}{3}\right)^{\prime}-\dfrac{1}{\sin ^2\left(\frac{\pi}{4}-x\right)} \cdot(-1)=$
$=-\cos ^2 \dfrac{x}{3} \cdot \sin \dfrac{x}{3}+\dfrac{1}{\sin ^2\left(x-\frac{\pi}{4}\right)}=\dfrac{1}{\sin ^2\left(x-\frac{\pi}{4}\right)}-\cos ^2 \dfrac{x}{3} \cdot \sin \dfrac{x}{3}$
$f^{\prime}\left(\dfrac{3 \pi}{4}\right)=\dfrac{1}{\sin ^2 \frac{\pi}{2}}-\cos ^2 \dfrac{\pi}{4} \sin \dfrac{\pi}{4}=1-\dfrac{\sqrt{2}}{4}$