d/dx(1+sinx/cosx)=cosx/dx хәл итегез

$$\frac { d } { d x } ( \frac { 1 + \sin x } { \cos x } ) = \frac { \cos x } { d x }$$

d өчен чишелеш

$$d=\frac{\left(\cos(x)\right)^{3}}{x\left(\sin(x)+1\right)}$$

$$x\neq 0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+\frac{\pi }{2}$$

Сызыкча тигезләмәне чишү адымнары

Үзгәртүчән d 0-гә тигез булырга мөмкин түгел, чөнки нольгә бүлү билгеләнмәгән. Тигезләмәнең ике ягын dx тапкырлагыз.

$$dx\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1+\sin(x)}{\cos(x)})=\cos(x)$$

Тигезләмә стандарт формасында.

$$x\left(-\frac{\left(\sin(x)+1\right)\frac{\mathrm{d}}{\mathrm{d}x}(\cos(x))}{\left(\cos(x)\right)^{2}}+\frac{\frac{\mathrm{d}}{\mathrm{d}x}(\sin(x))}{\cos(x)}\right)d=\cos(x)$$

Ике якны x\left(\frac{\mathrm{d}}{\mathrm{d}x}(\sin(x))\left(\cos(x)\right)^{-1}-\left(1+\sin(x)\right)\frac{\mathrm{d}}{\mathrm{d}x}(\cos(x))\left(\cos(x)\right)^{-2}\right)-га бүлегез.

$$\frac{x\left(-\frac{\left(\sin(x)+1\right)\frac{\mathrm{d}}{\mathrm{d}x}(\cos(x))}{\left(\cos(x)\right)^{2}}+\frac{\frac{\mathrm{d}}{\mathrm{d}x}(\sin(x))}{\cos(x)}\right)d}{x\left(-\frac{\left(\sin(x)+1\right)\frac{\mathrm{d}}{\mathrm{d}x}(\cos(x))}{\left(\cos(x)\right)^{2}}+\frac{\frac{\mathrm{d}}{\mathrm{d}x}(\sin(x))}{\cos(x)}\right)}=\frac{\cos(x)}{x\left(-\frac{\left(\sin(x)+1\right)\frac{\mathrm{d}}{\mathrm{d}x}(\cos(x))}{\left(\cos(x)\right)^{2}}+\frac{\frac{\mathrm{d}}{\mathrm{d}x}(\sin(x))}{\cos(x)}\right)}$$

x\left(\frac{\mathrm{d}}{\mathrm{d}x}(\sin(x))\left(\cos(x)\right)^{-1}-\left(1+\sin(x)\right)\frac{\mathrm{d}}{\mathrm{d}x}(\cos(x))\left(\cos(x)\right)^{-2}\right)'га бүлү x\left(\frac{\mathrm{d}}{\mathrm{d}x}(\sin(x))\left(\cos(x)\right)^{-1}-\left(1+\sin(x)\right)\frac{\mathrm{d}}{\mathrm{d}x}(\cos(x))\left(\cos(x)\right)^{-2}\right)'га тапкырлауны кире кага.

$$d=\frac{\cos(x)}{x\left(-\frac{\left(\sin(x)+1\right)\frac{\mathrm{d}}{\mathrm{d}x}(\cos(x))}{\left(\cos(x)\right)^{2}}+\frac{\frac{\mathrm{d}}{\mathrm{d}x}(\sin(x))}{\cos(x)}\right)}$$

\cos(x)'ны x\left(\frac{\mathrm{d}}{\mathrm{d}x}(\sin(x))\left(\cos(x)\right)^{-1}-\left(1+\sin(x)\right)\frac{\mathrm{d}}{\mathrm{d}x}(\cos(x))\left(\cos(x)\right)^{-2}\right)'га бүлегез.

$$d=\frac{\left(\cos(x)\right)^{3}}{x\left(\sin(x)+1\right)}$$

Үзгәртүчән d 0-гә тигез булырга мөмкин түгел.

$$d=\frac{\left(\cos(x)\right)^{3}}{x\left(\sin(x)+1\right)}\text{, }d\neq 0$$