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I-solve ang d
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dx\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1+\sin(x)}{\cos(x)})=\cos(x)
Ang variable d ay hindi katumbas ng 0 dahil hindi tukoy ang division by zero. I-multiply ang magkabilang dulo ng equation gamit ang dx.
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)
Ang equation ay nasa standard form.
\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)}
I-divide ang magkabilang dulo ng equation gamit ang 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{\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)}
Kapag na-divide gamit ang 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), ma-a-undo ang multiplication gamit ang 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)}
I-divide ang \cos(x) gamit ang 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)}\text{, }d\neq 0
Ang variable d ay hindi katumbas ng 0.