Question

Find derivatives y=tanh sqrt 1+t^2 ?

Answer 1

`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))`