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Exercise

dydx=tan(x+y−4)\frac{dy}{dx}=tan\left(x+y-4\right)

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Final answer to the exercise

ln⁡(−(tan⁡(x+y−42)−1)2+22)−12ln⁡(1+tan⁡(x+y−42)2)+arctan⁡(tan⁡(x+y−42))=x+C0\ln\left(\frac{\sqrt{-\left(\tan\left(\frac{x+y-4}{2}\right)-1\right)^2+2}}{\sqrt{2}}\right)-\frac{1}{2}\ln\left(1+\tan\left(\frac{x+y-4}{2}\right)^{2}\right)+\arctan\left(\tan\left(\frac{x+y-4}{2}\right)\right)=x+C_0

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dydx =tan(x+y−4)
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