d/dx(1+sinx/cosx)=cosx/dx सोडवा

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

d साठी सोडवा

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

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

रेखीय समीकरण सोडविण्यासाठी चरणे

शून्याने भागाकार करणे परिभाषित नसल्याने चल d हे 0 च्या समान असता कामा नये. समीकरणाच्या दोन्ही बाजूंना dx ने गुणाकार करा.

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

समीकरण मानक रूपामध्ये आहे.

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

दोन्ही बाजूंना 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) ने विभागा.

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

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) ने केलेला भागाकार 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) ने केलेला गुणाकार पूर्ववत करतो.

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

\cos(x) ला 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) ने भागा.

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

चल d हे 0 च्यास मान असता कामा नये.

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