y= (e^u - e^-u) / (e^u + e^-u)?

Question

find the second derivative using the chian rule. show steps please!

Answer 1

well,

(e^u - e^-u)/2 is sinh(u)
(e^u + e^-u)/2 is cosh(u)

sinh(u)/cosh(u) = [(e^u - e^-u)/2] / [(e^u + e^-u)/2]
sinh(u)/cosh(u) = (e^u - e^-u) / (e^u + e^-u) = tanh(u)

so y = tanh(u)

if u is just a v ariable and not some other function.
d[tanh(u)] = sech^2(u)
d[sech^2(u)] = 2 sech(u) * sech(u)*tan(u)

If u is some other function, yes you'd need to use chain rule, as well as product rule.
so y = tanh(u)
d[tanh(u)] = sech^2(u) * du/dx <-- first derivative
d[sech^2(u) * du/dx] = sech^2(u) * d[du/dx] + 2 sech(u) * sech(u)*tanh(u) * du/dx <-- second derivative