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Solving: ddx(y2−3sec⁡(x)tan⁡(x)+e5xcos⁡(y)=8)\frac{d}{dx}\left(y^2-3\sec\left(x\right)\tan\left(x\right)+e^{5x}\cos\left(y\right)=8\right)
Solve instead: dydx(y2−3sec⁡(x)tan⁡(x)+e5xcos⁡(y)=8)\frac{dy}{dx}\left(y^2-3\sec\left(x\right)\tan\left(x\right)+e^{5x}\cos\left(y\right)=8\right)

Exercise

dydx(y2−3sec⁡(x)tan⁡(x)+e5xcos⁡(y)=8)\frac{dy}{dx}\left(y^2-3\sec\left(x\right)\tan\left(x\right)+e^{5x}\cos\left(y\right)=8\right)

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Final answer to the exercise

2y⋅y′−3(sin⁡(x)2cos⁡(x)3+sec⁡(x)3)+5e5xcos⁡(y)−e5xy′sin⁡(y)=02y\cdot y^{\prime}-3\left(\frac{\sin\left(x\right)^2}{\cos\left(x\right)^{3}}+\sec\left(x\right)^{3}\right)+5e^{5x}\cos\left(y\right)-e^{5x}y^{\prime}\sin\left(y\right)=0

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dydx (y2−3sec(x)tan(x)+e5xcos(y)=8)
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