How do you differentiate #f(x)=sqrtcos(e^(4x))# using the chain rule.?

1 Answer
May 25, 2016

Answer:

#f'(x) = frac{-2sin(e^(4x)) e^(4x)}{sqrt(cos(e^(4x)))}#

Explanation:

Let

  • #p = 4x#
  • #q = e^p#
  • #r = cos(q)#
  • #s = sqrtr#

Therefore, to differentiate #f(x)#, we need to find #f'(x)#, which is the same as #frac{"d"s}{"d"x}#.

Applying the chain rule,

#frac{"d"s}{"d"x} = frac{"d"s}{"d"r} frac{"d"r}{"d"q} frac{"d"q}{"d"p} frac{"d"p}{"d"x}#

#= 1/(2sqrtr) * (-sin(q)) * e^p * (4)#

#= frac{-2sin(e^(4x)) e^(4x)}{sqrt(cos(e^(4x)))}#