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