सोडोवचें d/dx(1+sinx/cosx)=cosx/dx

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

d खातीर सोडोवचें

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

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

लिनियर समिकरण सोडोवंक पांवडे

विभागणी शुन्यची व्याख्या नाशिल्ल्यान अचल d हो 0 च्या समान आसूंक शकना. dx वरवीं समिकरणाच्या दोनूय कुशींक गुणाकार करचो.

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

समिकरण प्रमाणिक स्वरूपांत आसा.

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

दोनुय कुशींक 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) न भाग लावचो.

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

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) वरवीं भागाकार केल्यार 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) वरवीं केल्लो गुणाकार काडटा.

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

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) न\cos(x) क भाग लावचो.

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

अचल d हो 0 कडेन समान आसूंक शकना.

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