Given: #y = (cos(e^(x^5sinx)))^(1/2)#
Use the chain rule to differentiate.
Derivative rules needed:
Power rule: #" "(u^n)' = n u^(n-1)#
#(cos u)' = -u' sin u; " "(e^u)' = u' e^u#
product rule: #" "(uv)' = uv' + v u'#
Start out with the power rule where #u = cos(e^(x^5sinx)); n = 1/2#:
#y' = 1/2(cos(e^(x^5sinx)))^(-1/2) * (du)/(dx)(cos u)#
Let #u = e^(x^5sinx)#:
#y' = 1/2(cos(e^(x^5sinx)))^(-1/2) (-sin(e^(x^5 sinx)))* (du)/(dx)(e^u) #
Let #u = x^5sinx#:
#y' = 1/2(cos(e^(x^5sinx)))^(-1/2) (-sin(e^(x^5 sinx)))e^(x^5 sinx)* d/(dx) (x^5sinx)#
Use the product rule, #u = x^5, u' = 5x^4, v = sinx, v' = cos x#:
#y' = 1/2(cos(e^(x^5sinx)))^(-1/2) (-sin(e^(x^5 sinx)))e^(x^5 sinx) (x^5 cos x + 5 x^4 sin x)#
Rearrange and factor (GCF = #x^4#):
#y' = (-x^4e^(x^5sinx)sin(e^(x^5sinx))(xcosx + 5sinx))/(2sqrt(cos(e^(x^5sinx)))#