How do you differentiate #f(x) = x/sqrt(arctan(e^(x1)) # using the chain rule?
1 Answer
Answer:
Explanation:
We're asked to find the derivative
#d/(dx) [x/(sqrt(arctan[e^(x1)]))]#
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^(x1)]))# :
#= (d/(dx)[x])/(sqrt(tan^1[e^(x1)])) + xd/(dx)[1/(sqrt(tan^1[e^(x1)]))]#
The derivative of
#= 1/(sqrt(tan^1[e^(x1)])) + xd/(dx)[1/(sqrt(tan^1[e^(x1)]))]#
Now let's use the chain rule, which will be
#d/(dx) [1/(sqrt(tan^1[e^(x1)]))] = d/(du )[1/sqrtu] (du)/(dx)#
where

#u = tan^1[e^(x1)]# 
#d/(du) [1/sqrtu] = 1/(2u^(3"/"2))# :
#= 1/(sqrt(tan^1[e^(x1)])) + x((d/(dx)[tan^1[e^(x1)]])/(2tan^1[e^(x1)]^(3"/"2)))#
Now we'll use the chain rule again:
#d/(dx) [ tan^1[e^(x1)]] = d/(du) [tan^1u] (du)/(dx)#
where

#u = e^(x1)# 
#d/(du) [tan^1u] = 1/(1+u^2)# :
#= 1/(sqrt(tan^1[e^(x1)]))  ((d/(dx)[e^(x1)])/(e^(2x2)+1)) (x)/(2tan^1[e^(x1)]^(3"/"2))#
Finally, we'll use the chain rule a third time:
#d/(dx) [e^(x1)] = d/(du) [e^u] (du)/(dx)#
where

#u = x1# 
#d/(du) [e^u] = e^u# :
#= 1/(sqrt(tan^1[e^(x1)]))  e^(x1)d/(dx)[x1] (x)/(2(1+e^(2x2))tan^1[e^(x1)]^(3"/"2))#
The derivative of
#color(blue)(ulbar(stackrel(" ")(" "= 1/sqrt(tan^1[e^(x1)])  (xe^(x1))/(2(1+e^(2x2))tan^1[e^(x1)]^(3"/"2)))" "))#