What is the equation of the tangent line to function y=e^x/x at point (1, e).?

Question

I get the slope of the tangent is 0 and the equation of the tangent line is y=e but don't think its right.

Answer 1

The slope of the tangent line is the derivative of the given function at x =1. Differentiating, according to the rule for differentiating a quociente of 2 differentiable functions, we get

y' = (x e^x - e^x)/(x^2) = (e^x(x -1))/(x^2). So, the slope is indeed = e(1 -1)/1 = 0. And since the tangent line is, therefore, horizontal and passes through (1, e), it's equation is indeed y =e. y has a global minimum on (0, oo) at x =e.

Your answer is right.

Answer 2

so you take the derivative of the function and evaluate at x=1 to get slope which you already did.
then plug the values into point slope form:
y-y1=m(x-x1)

y-e=0(x-1)

y=e you were correct

Answer 3

i think you're right

y=e^x/x

dy = { d(e^x) x - e^x d(x) } / x^2
= { e^x x - e^x } / x^2

tangent line at (1,e)
dy (1) = { e^1 1 - e^1 } / 1^2 = 0

Answer 4

y' = ( xe^x - e^x )/ x^2 = (x-1)e^x/x^2
at (1,e) y'=0

So y -e =0 => y=e