What is the equation of the tangent line of #f(x) =e^xlnx-x# at #x=2#?

1 Answer
Jan 20, 2016

For finding the equation of the tangent of the line, we need need to find the slope of the line first.
Slope of the line at a point can be found by differentiating the function, so
#frac{d}{dx}(f(x))=\frac{d}{dx}(e^xln(x)-x)=ln(x)\frac{d}{dx}(e^x)+e^x\frac{d}{dx}(ln(x))-\frac{dx}{dx}#
So, #m=e^xln(x)+e^x/x-1#

At, #x=2 " " m=e^2ln2+e^2/2-1#
Also, at #x=2# #f(2)=e^2ln2-2=y#
Given slope equation is #y-y_o=m(x-x_o)#

Substituting all we got,
#y-e^2ln2+2=(e^2ln2+e^2-1)(x-2)#

Simplification is left as exercise for the reader.