Answer
$y^{\prime}=-10x\cdot \mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}(5x^{2})\tanh(5x^{2})$
Work Step by Step
Apply Th. 5.18:
Let $u$ be a differentiable function of $x$.
$ \displaystyle \frac{d}{dx} [{\rm sech} u] =-({\rm sech} u\tanh u)u^{\prime}$
---
$\displaystyle \frac{d}{dx}=\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}(5x^{2})$
$=-\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}(5x^{2})\tanh(5x^{2})(10x)$
$=-10x\cdot \mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}(5x^{2})\tanh(5x^{2})$
