How do you differentiate #f(x)=lnsqrt(-e^(4x)-2)# using the chain rule.?

1 Answer
Mar 13, 2018

Answer:

#(2e^(4x))/(e^(4x)+2)#

Explanation:

According to the chain rule, #(f(g(x)))'=f'(g(x))*g'(x)#.

We have #f(u)=lnu# where #u=g(x)=sqrt(-e^(4x)-2)#

We need to find #d/(du)lnu*d/(dx)sqrt(-e^(4x)-2)#

Since #d/dxlnx=1/x#, we write:

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

Concentrating on the bracket, let's define two other functions:

#h(v)=sqrt(v)# where #v=j(x)=-e^(4x)-2#

Since #d/dxsqrt(x)=1/(2sqrt(x))#, we write:

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

Concentrating on the bracket, we define two more functions:

#k(z)=-e^z-2#, where #z=l(x)=4x#

We have #d/(dz)(-e^z-2)*d/dx(4x)#

#-(d/(dz)e^z+d/(dz)2)*4d/dxx#

#-(e^z)*4#

#-4e^z#

Overall, we have:

#1/u(1/(2sqrt(v))(-4e^z))#

Well, that's not right, right? We just have a bunch of nonsense variables. But remember that:

#u=sqrt(-e^(4x)-2)#

#v=-e^(4x)-2#

#z=4x#

We can input our stuff in:

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

And simplify:

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

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

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

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

#(2e^(4x))/(e^(4x)+2)#

The answer.