How do you find the derivative of #e^ [2 tan(sqrt x)]#?

1 Answer
Aug 11, 2016

You do it one step at a time, keeping in mind the chain rule.

Any time you take the derivative of a function that contains a nested function (otherwise known as a composite function), take the derivative of the nested function as well.

That is, #d/(dx)[f(u(x))] = (df(u))/(du(x))*(du(x))/(dx)#. So:

  • The derivative of #f(u) = e^u# is #e^u ((du)/(dx))#.
  • The derivative of #f(u) = tanu# is #sec^2u ((du)/(dx))#
  • The derivative of #f(x) = sqrtx# is #1/(2sqrtx)#.

Therefore:

#color(blue)(d/(dx)[e^(2tansqrtx)])#

#= e^(2tansqrtx) * stackrel("Chain Rule")overbrace(2d/(dx)[tansqrtx])#

Here, #u = tansqrtx#, so #color(green)(d/(dx)[u(x)] = sec^2sqrtx * d/(dx)[sqrtx])#:

#=> e^(2tansqrtx) * 2(sec^2sqrtx * stackrel("Chain Rule Again")overbrace(d/(dx)[sqrtx]))#

#= cancel(2)e^(2tansqrtx)sec^2sqrtx * 1/(cancel(2)sqrtx)#

And here, #u = sqrtx#, so now #color(green)((du)/(dx) = 1/(2sqrtx))#:

#=> color(blue)((e^(2tansqrtx)sec^2sqrtx)/sqrtx)#

That's as simple an answer as it gets, so don't be surprised if you get this. :-)