To find the equation of the line, we need a point the line passes through and a gradient.
As we wish to find the line that is tangent to #x=2# then we can use #f(2)# to find a point where the line passes through:
#f(x)=xe^sqrt(x) -> f(2) = 2e^sqrt(2)#
So we know at least the line passes through #(2, 2e^sqrt(2))#
Now we need the gradient which we can obtain from #f'(x)# the derivative.
Differentiate #f(x)# using the product rule and chain rule to get:
#f'(x) = e^sqrt(x)+ x/(2sqrt(x))e^sqrt(x)=e^sqrt(x)(1+sqrt(x)/2)#
We have simplified the function by factoring out #e^sqrt(x)# and dividing the #x/(2sqrt(x))# term by #sqrt(x)#. We can now obtain the gradient:
#f'(2) = e^sqrt(2)(1+sqrt(2)/2)=e^sqrt(2)(1+1/(sqrt(2)))=m#
We can now put this into the equation of a line:
#y-b = m(x-a)# where #(a,b)# is our know point from above. That gives us:
#y-2e^sqrt(2) = e^sqrt(2)(1+1/sqrt(2))(x-2)#
We can the #e^sqrt(2)# over to the right by adding then factoring to simplify and get:
#y = e^sqrt(2)[(1+1/sqrt(2))(x-2)+2]#
