Find derivatives y=tanh sqrt 1+t^2 ?
1 Answer
Jun 2, 2018
# dy/d = ( t \ sech^2 sqrt(1+t^2))/(sqrt(1+t^2)) #
Explanation:
We seek:
# dy/dt# , where#y =tanh sqrt(1+t^2) #
Using the known result:
#d/dt tanh = sech^2t#
In conjunction with the chain rule, we get:
# dy/dt = (sech^2 sqrt(1+t^2))(d/dt sqrt(1+t^2)) #
# \ \ \ \ \ = sech^2 sqrt(1+t^2) \ (1/2(1+t^2)^(-1/2)d/dt(1+t^2)) #
# \ \ \ \ \ = sech^2 sqrt(1+t^2) \ (1/(2sqrt(1+t^2)))(2t) #
# \ \ \ \ \ = ( t \ sech^2 sqrt(1+t^2))/(sqrt(1+t^2)) #
