#lny = ln(lnx)^cosx#
#lny = cosxln(lnx)#
Differentiate #ln(lnx)# using the chain rule.
Let #y = lnu# and #u = lnx#.
#dy/(du) = 1/u# and #(du)/dx = 1/x#
#dy/dx = dy/(du) xx (du)/dx#
#dy/dx= 1/u xx 1/x#
#dy/dx = 1/lnx xx 1/x#
#dy/dx = 1/(xlnx)#
We can differentiate the entire function using the product rule.
#1/y(dy/dx) = -sinxln(lnx) + cosx/(xlnx)#
#dy/dx = y(-sinxln(lnx) + cosx/(xlnx))#
#dy/dx= (lnx)^cosx(-sinxln(lnx) + cosx/(xlnx))#
Now, determine the slope of the tangent by inserting your point #x= a# within the derivative.
#m_"tangent" = (lne)^cos(e)(-sin(e)ln(lne) + cose/(elne))#
#m_"tangent" = 1(cos(e))/e#
#m_"tangent" = cose/e#
We now determine the equation of the tangent line.
#y - y_1 = m(x- x_1)#
#y - 1 = cose/e(x - e)#
#y = cose/ex - cose +1#
Hopefully this helps!
