Hi there!

I was doing some homework tasks for math, and I got stuck on 10ii, proving that the stationary point is a maximum. I was wondering how would I be able to do this question. Would I need to second derivative at all?

Thank you!!!

The question wants you to avoid using the second derivative but rather "consider the gradient on each side of \(x=e\)".

\[ \text{So we just sub some numeric values in.}\\ \text{For a point on the left, we can try }x=2.\\ \text{When }x=2\text{, the calculator says that }\frac{dy}{dx} = 0.0767\dots > 0 \]

Therefore on the left of \(x=e\), the curve is increasing.

\[ \text{For a point on the right, we can try }x=3.\\ \text{When }x=3\text{, the calculator says that }\frac{dy}{dx} = -0.0109\dots < 0\]

Therefore on the right of \(x=e\), the curve is decreasing.

\[ \text{So the curve increases, until it reaches }x=e.\\ \text{Then after }x=e\text{, it decreases.}\\ \text{That's all that is required to conclude that }x=e\\ \text{must therefore be a local maximum.} \]

This method (which is really just a table of values) should be noted, because sometimes computing the second derivative is just nastily exhausting effort. The whole point of this method is to get you out of that trouble.

Remark: Second derivative

\[ \text{If you insist on using the second derivative}\\ \text{you more or less need the quotient rule }\textit{again}. \]

\begin{align*}\frac{dy}{dx} &= \frac{x\cdot \frac1x - 1 \cdot \ln x}{x^2}\\ &= \frac{1-\ln x}{x^2}\\ \therefore \frac{d^2y}{dx^2} &= \frac{x^2\cdot -\frac1x - 2x(1-\ln x)}{x^4}\end{align*}

\[ \text{At this point, you can now sub }x=e\text{ in straight away.}\]

This can optionally be simplified to \( \frac{d^2y}{dx^2} = \frac{-3x+2x\ln x}{x^4} = \frac{-3+2\ln x}{x^3}\) but you don't really need to. You can just sub \(x=e\) into the unsimplified expression.

For this question though, in theory you should not do this. This is because they told you to use the other method.