Here we shall make use of both the quotient rule and the chain rule. The quotient states that for any expression that can be written as #f (x)/g (x)#, , where f and get are differentiable with respect to x and g is not 0, the derivative can be found using the following formula:
#d/(dx) (f (x)/g (x)) = (f'(x)g (x)-f (x)g'(x))/((g (x))^2#
The chain rule states that when given an expression #u (v (x))#, the derivative takes the form #(d/(dx)v (x))*d/(dv)u(v (x))#
Above, we must use the chain rule on the numerator. Here u is #e^v# and v is #tan x# . The derivative of #e^v# with respect to v is simply #e^v#, and the derivative of tan (x) with respect to X is simply #sec^2 (x)# (recall the quotient rule, and that tan = sin/cos, and the Riva times of sin and cos (cos and -sin respectively), thus #d/d(dx)(sin (x)/cos (x)) = (cos^2 (x) + sin^2 (x))/cos^2 (x) 1/(cos^2 (x)) = sec^2 (x)#
Further, the derivative of #(sqrt x = x^(1/2)# is simply #x^-(1/2)/2#. Thus, we have #f'(x) = sec^2 (x)*e^(tan x)# and #g'(x) = x^(-1/2)/2#. Therefore, the quotient rule gives us...
#d/(dx) (e^(tan x)/x^(1/2)) = (sec^2 (x)e^tan (x)x^(1/2) - (e^tan (x)x^-(1/2))/2)/x# , which for any nonzero X is #sec^2 (x)e^tan (x)x^(-1/2)- (e^tan (x)x^-(3/2))/2#
This may simplify further, so I invite anyone who can manage it to do so.
