How do you differentiate #f(x) = x/sqrt(arctan(e^(x-1)) # using the chain rule?

1 Answer
Aug 12, 2017

Answer:

#d/(dx) [x/(sqrt(arctan[e^(x-1)]))] = color(blue)(1/sqrt(tan^-1[e^(x-1)]) - (xe^(x-1))/(2(1+e^(2x-2))tan^-1[e^(x-1)]^(3"/"2)))#

Explanation:

We're asked to find the derivative

#d/(dx) [x/(sqrt(arctan[e^(x-1)]))]#

Let's first use the product rule, which is

#d/(dx) [uv] = v(du)/(dx) + u(dv)/(dx)#

where

  • #u = x#

  • #v = 1/(sqrt(arctan[e^(x-1)]))#:

#= (d/(dx)[x])/(sqrt(tan^-1[e^(x-1)])) + xd/(dx)[1/(sqrt(tan^-1[e^(x-1)]))]#

The derivative of #x# is #1#:

#= 1/(sqrt(tan^-1[e^(x-1)])) + xd/(dx)[1/(sqrt(tan^-1[e^(x-1)]))]#

Now let's use the chain rule, which will be

#d/(dx) [1/(sqrt(tan^-1[e^(x-1)]))] = d/(du )[1/sqrtu] (du)/(dx)#

where

  • #u = tan^-1[e^(x-1)]#

  • #d/(du) [1/sqrtu] = -1/(2u^(3"/"2))#:

#= 1/(sqrt(tan^-1[e^(x-1)])) + x(-(d/(dx)[tan^-1[e^(x-1)]])/(2tan^-1[e^(x-1)]^(3"/"2)))#

Now we'll use the chain rule again:

#d/(dx) [ tan^-1[e^(x-1)]] = d/(du) [tan^-1u] (du)/(dx)#

where

  • #u = e^(x-1)#

  • #d/(du) [tan^-1u] = 1/(1+u^2)#:

#= 1/(sqrt(tan^-1[e^(x-1)])) - ((d/(dx)[e^(x-1)])/(e^(2x-2)+1)) (x)/(2tan^-1[e^(x-1)]^(3"/"2))#

Finally, we'll use the chain rule a third time:

#d/(dx) [e^(x-1)] = d/(du) [e^u] (du)/(dx)#

where

  • #u = x-1#

  • #d/(du) [e^u] = e^u#:

#= 1/(sqrt(tan^-1[e^(x-1)])) - e^(x-1)d/(dx)[x-1] (x)/(2(1+e^(2x-2))tan^-1[e^(x-1)]^(3"/"2))#

The derivative of #x-1# is #1#:

#color(blue)(ulbar(|stackrel(" ")(" "= 1/sqrt(tan^-1[e^(x-1)]) - (xe^(x-1))/(2(1+e^(2x-2))tan^-1[e^(x-1)]^(3"/"2)))" ")|)#