Question

How do you use the first and second derivatives to sketch `y=e^x/x`?

Answer 1

graph{(e^x)/x [-22, 18, -9.88, 10.12]}

Explanation:

`y=e^x/x`

to sketch the graph, we need to examine various behaviour of the function

roots:
`y =0 => e^x/x=0`
`:. e^x=0`
But as `e^x>0 AA x in RR =>` no roots

behaviour as `x->+-oo`
As `x->-oo=>y->e^-oo/-oo->0^-`
As `x->oo=>y->e^oo/oo->oo`

asymptotes

denominator`=0 => x=0`
so a vertical asymptote when `x=0`

turning (or critical points)
`y=e^x/x`, so differentiating wrt `x` using the quotient rule
`d/dx(u/v)=(v(du)/dx - u(dv)/dx)/v^2`
gives:
`dy/dx = (xd/dx(e^x)-e^xd/dx(x))/x^2`
`= (xe^x-e^x)/x^2`
`= ((x-1)e^x)/x^2`

At a critical point `dy/dx=0 => ((x-1)e^x)/x^2 = 0`
`:. (x-1)e^x=0`

`:. e^x=0` (no solutions); or `x-1=0=>x=1`

When `x=1 => y=e^1/1=e` (or `~~2.7` )
so there is a critical point at `(1,e)`

Nature of the critical points:
We need to look at the second derivative;
we can rearrange
`dy/dx = ((x-1)e^x)/x^2` as
`dy/dx = e^x/x-e^e/x^2`
we already know the derivative of `e^x/x` (it is the above); so we must use the quotient rule again to find the derivative of `e^x/x^2`

so we get
`(d^2y)/dx^2 = ((x-1)e^x)/x^2-d/dx(e^x/x^2)`
`:. (d^2y)/dx^2 = ((x-1)e^x)/x^2 - {(x^2d/dx(e^x) - e^xd/dx(x^2))/(x^2)^2}`
`:. (d^2y)/dx^2 = ((x-1)e^x)/x^2 - {(x^2e^x - 2xe^x)/(x^4)}`

That's quite a complex expression, so lets not even bother trying to simplify any more as we increase the chance of making a mistake; just substitute `x=1` to determine the nature of the turning point;

`x=1 => (d^2y)/dx^2 = 0 - {(e^1 - 2e^1)/(1}}=e>0 =>` minimum

Sketching the graph:
We now have enough to sketch the graph which actually looks like this;
graph{(e^x)/x [-22, 18, -9.88, 10.12]}