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Solving: ddx(y=tanh(x)2ln⁡(sech(y)))\frac{d}{dx}\left(y=\mathrm{tanh}\left(x\right)^2\ln\left(\mathrm{sech}\left(y\right)\right)\right)
Solve instead: dydx(y=ln(sech)tanh2x)\frac{dy}{dx}\left(y=ln\left(sech\right)tanh^2x\right)

Exercise

dydx(y=ln(sech)tanh2x)\frac{dy}{dx}\left(y=ln\left(sech\right)tanh^2x\right)

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Final answer to the exercise

y′=2sech(x)2ln⁡(sech(y))tanh(x)1+tanh(x)2tanh(y)y^{\prime}=\frac{2\mathrm{sech}\left(x\right)^2\ln\left(\mathrm{sech}\left(y\right)\right)\mathrm{tanh}\left(x\right)}{1+\mathrm{tanh}\left(x\right)^2\mathrm{tanh}\left(y\right)}

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dydx (y=ln(sech)tanh2x)
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