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202112021444 Solution to 1979-CE-AMATH-I-1

Let \displaystyle{y=x+\frac{1}{x^2}}. Find \displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}} from first principles.

Solution.

From first principles,

\begin{aligned} y:=f(x) & =x+\frac{1}{x^2} \\ f(x+\Delta x) & =(x+\Delta x) + \frac{1}{(x+\Delta x)^2} \\ f(x+\Delta x)-f(x) & = \Delta x + \frac{1}{(x+\Delta x)^2} - \frac{1}{x^2} \\ & = \Delta x + \frac{x^2-(x+\Delta x)^2}{x^2(x+\Delta x)^2} \\ & = \Delta x - \frac{2x\Delta x+(\Delta x)^2}{x^2(x+\Delta x)^2} \\ \frac{f(x+\Delta x)-f(x)}{\Delta x} & = 1 - \frac{2x + \Delta x}{x^2(x+\Delta x)^2} \\ \lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} & = 1 - \frac{2x + 0}{x^2(x+0)^2} \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = 1 - \frac{2}{x^3} \\ \end{aligned}

Playground.

To make fun of calculus, do let

\begin{aligned} y & =x+\frac{1}{x^2} \\ x^2y & =x^3+1 \\ 0 & = x^3-x^2y+1 := f(x,y) \\ \end{aligned}

where f(x,y) is a degree 3 polynomial in two variables x and y.

The set of solutions to f(x,y)=0 is

\begin{cases} \enspace x \neq 0 \\ \enspace y = \frac{x^3+1}{x^2} \end{cases}

Computing \partial_xf and \partial_yf as follow:

\begin{aligned} \partial_xf & = \frac{\partial}{\partial x}f(x,y) \\ & = 3x^2-2xy \\ \end{aligned}

and

\begin{aligned} \partial_yf & = \frac{\partial}{\partial y}f(x,y) \\ & = -x^2 \\ \end{aligned},

we oversee the relation

\mathrm{d}f = (\partial_xf)(\Delta x)+(\partial_yf)(\Delta y).

Definition. (Singularity and Smoothness)

A point p=(a,b) on a curve \mathcal{C}=\{ (x,y)\in\mathbb{C}^2:f(x,y)=0\} is said to be singular if

\begin{aligned} \frac{\partial f}{\partial x}(a,b) & = 0 \\ \frac{\partial f}{\partial y}(a,b) & = 0 \\ \end{aligned}.

A point that is not singular is called smooth. If there is at least one singular point on \mathcal{C}, then curve \mathcal{C} is called a singular curve. If there are no singular points on \mathcal{C}, the curve \mathcal{C} is called a smooth curve.

C.f. Definition 1.9.1, T. Carrity, et al., Algebraic Geometry: A Problem Solving Approach

That said, in our scenario

f(x,y)=x^3-x^2y+1,

the set of candidates for singularity

\{ (a,b)\in\mathbb{C}^2\enspace \mathrm{s.t.}\enspace f(a,b)=0\textrm{ and }\partial_xf(a,b)=\partial_yf(a,b)=0\}.

is empty.

\therefore The curve \mathcal{C}=\{ (x,y)\in\mathbb{C}^2:x^3-x^2y+1=0\} is smooth, thus everywhere differentiable.

Wait, the function in question should be the curve

F(x)=\displaystyle{x+\frac{1}{x^2}};

still, for better or worse, F(x) is \textrm{\scriptsize{NOT}} a differentiable function because the derivative does \textrm{\scriptsize{NOT}} exist at x=0, for that point is a discontinuity.