Question

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

Answer 1

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.