# Is there any point (x, y) on the curve y=x^(x(1+1/y)), x > 0, at which the tangent is parallel to the x-axis?

Aug 20, 2016

There is no such point, as far as my math goes.

#### Explanation:

First, let's consider the conditions of the tangent if it is parallel to the $x$-axis. Since the $x$-axis is horizontal, any line parallel to it must also be horizontal; so it follows that the tangent line is horizontal. And, of course, horizontal tangents occur when the derivative equals $0$.

Therefore, we must first start by finding the derivative of this monstrous equation, which can be accomplished through implicit differentiation:
$y = {x}^{x + \frac{x}{y}}$
$\to \ln y = \left(x + \frac{x}{y}\right) \ln x$

Using the sum rule, chain rule, product rule, quotient rule, and algebra, we have:
$\frac{d}{\mathrm{dx}} \left(\ln y\right) = \frac{d}{\mathrm{dx}} \left(\left(x + \frac{x}{y}\right) \ln x\right)$
$\to \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{1}{y} = \left(x + \frac{x}{y}\right) ' \left(\ln x\right) + \left(x + \frac{x}{y}\right) \left(\ln x\right) '$
$\to \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{1}{y} = \left(x + \frac{x}{y}\right) ' \left(\ln x\right) + \left(x + \frac{x}{y}\right) \left(\ln x\right) '$
$\to \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{1}{y} = \left(1 + \frac{x ' y - x \frac{\mathrm{dy}}{\mathrm{dx}}}{y} ^ 2\right) \left(\ln x\right) + \left(x + \frac{x}{y}\right) \left(\frac{1}{x}\right)$
$\to \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{1}{y} = \ln x + \ln x \left(\frac{y - x \frac{\mathrm{dy}}{\mathrm{dx}}}{y} ^ 2\right) + 1 + \frac{1}{y}$
$\to \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{1}{y} = \ln x + \ln x \left(\frac{1}{y} - \frac{x \frac{\mathrm{dy}}{\mathrm{dx}}}{y} ^ 2\right) + 1 + \frac{1}{y}$
$\to \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{1}{y} = \ln x + \frac{\ln x}{y} - \frac{x \ln x \frac{\mathrm{dy}}{\mathrm{dx}}}{y} ^ 2 + 1 + \frac{1}{y}$
$\to \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{1}{y} + \frac{x \ln x \frac{\mathrm{dy}}{\mathrm{dx}}}{y} ^ 2 = \ln x + \frac{\ln x}{y} + 1 + \frac{1}{y}$
$\to \frac{\mathrm{dy}}{\mathrm{dx}} \left(\frac{1}{y} + \frac{x \ln x}{y} ^ 2\right) = \ln x + \frac{\ln x}{y} + 1 + \frac{1}{y}$
$\to \frac{\mathrm{dy}}{\mathrm{dx}} \left(\frac{y + x \ln x}{y} ^ 2\right) = \ln x + \frac{\ln x}{y} + 1 + \frac{1}{y}$
$\to \frac{\mathrm{dy}}{\mathrm{dx}} \left(\frac{y + x \ln x}{y} ^ 2\right) = \frac{y \ln x + \ln x + 1 + y}{y}$
$\to \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{y \ln x + \ln x + 1 + y}{y}}{\frac{y + x \ln x}{y} ^ 2}$
$\to \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \left(y \ln x + \ln x + 1 + y\right)}{y + x \ln x}$

Wow...that was intense. Now we set the derivative equal to $0$ and see what happens.
$0 = \frac{y \left(y \ln x + \ln x + 1 + y\right)}{y + x \ln x}$
$0 = y \ln x + \ln x + 1 + y$
$- y \ln x - y = \ln x + 1$
$- y \left(\ln x + 1\right) = \ln x + 1$
$y \left(\ln x + 1\right) = - \left(\ln x + 1\right)$
$y = \frac{- \left(\ln x + 1\right)}{\ln x + 1}$
$y = - 1$

Interesting. Now let's plug in $y = - 1$ and see what we get for $x$:
$y = {x}^{x \left(1 + \frac{1}{y}\right)}$
$- 1 = {x}^{x \left(1 + \frac{1}{-} 1\right)}$
$- 1 = {x}^{x \left(1 - 1\right)}$
$- 1 = {x}^{0}$
$- 1 = 1$

Since this is a contradiction, we conclude that there are no points meeting this condition.

Aug 22, 2016

There not exists such a tangent.

#### Explanation:

$y = {x}^{x \left(1 + \frac{1}{y}\right)} \equiv {y}^{\frac{y}{y + 1}} = {x}^{x}$. Now calling $f \left(x , y\right) = {x}^{x} - {y}^{\frac{y}{y + 1}} = u \left(x\right) + v \left(y\right) = 0$ we have

$\mathrm{df} = {f}_{x} \mathrm{dx} + {f}_{y} \mathrm{dy} = \frac{\partial u}{\partial x} \mathrm{dx} + \frac{\partial v}{\partial y} \mathrm{dy} = 0$ then

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\frac{\partial u}{\partial x}}{\frac{\partial v}{\partial y}} = \frac{{x}^{x} \left(1 + L o {g}_{e} \left(x\right)\right) {\left(1 + y\right)}^{2}}{{y}^{\frac{y}{1 + y}} \left(1 + y + L o {g}_{e} \left(y\right)\right)} = \frac{\left(1 + L o {g}_{e} \left(x\right)\right) {\left(1 + y\right)}^{2}}{1 + y + L o {g}_{e} \left(y\right)}$

We see that $\frac{\mathrm{dy}}{\mathrm{dx}} = 0 \to \left\{{y}_{0} = - 1 , {x}_{0} = {e}^{- 1}\right\}$ but those values must verify:

$f \left(x , {y}_{0}\right) = 0$ and
$f \left({x}_{0} , y\right) = 0$

In the first case, ${y}_{0} = 1$ we have

${x}^{x} = - 1$ which is not attainable in the real domain.

In the second case, ${x}_{0} = {e}^{- 1}$ we have

${y}^{\frac{y}{y + 1}} = {e}^{- 1}$ or
$\frac{y}{y + 1} {\log}_{e} y = - 1$

but

$\frac{y}{y + 1} {\log}_{e} y > - 1$ so no real solution also.

Concluding, there is not such a tangent.

Aug 23, 2016

The answer from Dr, Cawa K, x = 1/e, is precise.

#### Explanation:

I had proposed this question to get this value precisely. Thanks to

Dr, Cawas for a decisive answer that approves the revelation that

the double precision y' remains 0 around this interval. y is

continuous and differentiable at x = 1/e. As both the 17-sd double

precision y and y' are 0, in this interval around x = 1/e, it was a

conjecture that x-axis touches the graph in between. And now, it is

proved. I think that the touch is transcendental. .