If #f(x)= csc7 x # and #g(x) = e^(1 +3x ) #, how do you differentiate #f(g(x)) # using the chain rule?

Question

If #f(x)=  csc7   x # and #g(x)  = e^(1  +3x   )  #, how do you differentiate #f(g(x))  # using the chain rule?

Luke Alvarez · Accepted Answer

#d/(dx)f(g(x))=-21e^((1+3x))csc(7e^(1+3x))cot(7e^(1+3x))# As #f(x)=csc7x# and #g(x)=e^(1+3x)# #f(g(x))=csc(7e^(1+3x))# The chain in the derivative of #f(g(x))# is in fact longer. We have #f(x)=csc(u(x))#, #u(g(x))=7g(x)#, #g(x)=e^(v(x))# and #v(x)=1+3x#. Hence #d/(dx)f(g(x))=(df)/(du)xx(du)/(dg)xx(dg)/(dv)xx(dv)/(dx)# = #-cscucotuxx7xxe^vxx3# = #-21e^((1+3x))csc(7e^(1+3x))cot(7e^(1+3x))#

Axel Alvarez · Answer

undefined To differentiate $ f(g(x)) $ using the chain rule, you follow these steps:

1. Compute the derivative of the outer function with respect to its inner function.
2. Multiply by the derivative of the inner function with respect to $ x $.

Applying this to $ f(g(x)) $, where $ f(x) = \csc(7x) $ and $ g(x) = e^{1 + 3x} $, the derivative is:

$$ \frac{d}{dx}\left( \csc(7e^{1 + 3x}) \right) = -7\csc(7e^{1 + 3x})\cot(7e^{1 + 3x})\cdot 3e^{1 + 3x} $$