Exercise
$\frac{d}{dx}\left(y=\frac{\cosh\left(x\right)}{1+\sech\left(x\right)}+\log\left(\sech\left(x\right)+\sech\left(3x^2+1\right)\right)\right)$
Step-by-step Solution
Final answer to the exercise
$y^{\prime}=\frac{\left(1+\mathrm{sech}\left(x\right)\right)\mathrm{sinh}\left(x\right)+\mathrm{sinh}\left(x\right)\mathrm{sech}\left(x\right)}{\left(1+\mathrm{sech}\left(x\right)\right)^2}+\frac{-\mathrm{sech}\left(x\right)\mathrm{tanh}\left(x\right)-6x\mathrm{sech}\left(3x^2+1\right)\mathrm{tanh}\left(3x^2+1\right)}{\ln\left(10\right)\left(\mathrm{sech}\left(x\right)+\mathrm{sech}\left(3x^2+1\right)\right)}$