Lahenda d/dx(1+sinx/cosx)=cosx/dx

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

Lahendage ja leidke d

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

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

Juhised lineaarvõrrandi lahendamiseks

Muutuja d ei tohi võrduda väärtusega 0, kuna nulliga jagamist pole määratletud. Korrutage võrrandi mõlemad pooled dx-ga.

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

Võrrand on standardkujul.

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

Jagage mõlemad pooled 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)-ga.

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

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)-ga jagamine võtab 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)-ga korrutamise tagasi.

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

Jagage \cos(x) väärtusega 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).

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

Muutuja d ei tohi võrduda väärtusega 0.

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