# 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)) #