What is the equation of the tangent line of f(x)=exln⁡x−x at x=2?

Question

Charlie Haley · Accepted Answer

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}{d x} (f (x)) = \frac{d}{d x} (e^{x} \ln (x) - x) = \ln (x) \frac{d}{d x} (e^{x}) + e^{x} \frac{d}{d x} (\ln (x)) - \frac{d x}{d x}

So,

m = e^{x} \ln (x) + \frac{e^{x}}{x} - 1

At,

x = 2 m = e^{2} \ln 2 + \frac{e^{2}}{2} - 1

Also, at

x = 2 f (2) = e^{2} \ln 2 - 2 = y

Given slope equation is

y - y_{0} = m (x - x_{0})

Substituting all we got,

y - e^{2} \ln 2 + 2 = (e^{2} \ln 2 + e^{2} - 1) (x - 2)

Simplification is left as exercise for the reader.

What is the equation of the tangent line

Answered question

Answer & Explanation

New Questions in Calculus 1