What is the derivative of this function #y=sec^-1(e^(2x))#?

2 Answers
Feb 19, 2018

#(2)/(sqrt(e^(4x)-1)#

Explanation:

As if #y=sec^-1x# the derivative is equel to #1/(xsqrt(x^2-1))#
so by using this formula and if #y=e^(2x)# then derivative is #2e^(2x)# so by using this relation in the formula we get the required answer. as #e^(2x)# is a function other than #x# that is why we need further derivative of #e^(2x)#

Feb 19, 2018

#2/(sqrt(e^(4x)-1))#

Explanation:

We have #d/dxsec^-1(e^(2x))#.

We can apply the chain rule, which states that for a function #f(u)#, its derivative is #(df)/(du)*(du)/dx#.

Here, #f=sec^-1(u)#, and #u=e^(2x)#.

#d/dxsec^-1(u)=1/(sqrt(u^2)sqrt(u^2-1))#. This is a common derivative.

#d/dxe^(2x)#. Chain rule again, here #f=e^u# and #x=2x#. The derivative of #e^u# is #e^u#, and the derivative of #2x# is #2#.

But here, #u=2x#, and so we finally have #2e^(2x)#.

So #d/dxe^(2x)=2e^(2x)#.

Now we have:

#(2e^(2x))/(sqrt(u^2)sqrt(u^2-1))#, but since #u=e^(2x)#, we have:

#(2e^(2x))/(sqrt((e^(2x))^2)sqrt((e^(2x))^2-1))#

#(2e^(2x))/(e^(2x)sqrt((e^(4x))-1))#

#2/(sqrt(e^(4x)-1))#, our derivative.