## 1. `$y = \frac{1}{x}$`

#

The derivative of `$y = \frac{1}{x}$`

can be computed in the following way.

First, I highly recommend you to watch this clip , where 3blue1brown visualizes this function.

The key part of the proof is that Since A and B are both on the curve of `$y = \frac{1}{x}$`

, the x coordinate times the y coordinate is 1. That is to say: `$A_x \cdot A_y = 1$`

and `$B_x \cdot B_y = 1$`

. If you know this, then you’ll know that the areas of the two shaded areas are equal. Therefore,

$$dx \cdot (\frac{1}{x} - df) = x \cdot df$$

We have:

$$x\cdot df = \frac{dx}{x} - dx\cdot df$$

So:

$$x = \frac{dx}{x \cdot df} - dx$$

Because `$dx$`

is extremely small, we can safely ignore it, and therefore, we have:

$$x = \frac{dx}{x \cdot df}$$

Multiply the above equation by `$x$`

and we have:

$$x^2 = \frac{dx}{df}$$

So we have:

$$\frac{df}{dx} = \frac{1}{x^2} = x^{-2}$$

Because `$df$`

is negative, the derivative should be negative as well, so:

$$\frac{df}{dx} = - \frac{1}{x^2} = - x^{-2}$$

This method is inspired by F J .

## 2. `$y = \sqrt{x}$`

#

It can be proven in two ways.

### Intuitive way #

First, let’s use the way suggested by 3blue1brown.

I’ll write `$d\sqrt{x}$`

as `$dy$`

.

We have

$$dx = 2\sqrt{x}\cdot dy + (dy)^2$$

Divide the equation by `$dy$`

and we have:

$$2\sqrt{x} + dy = \frac{dx}{dy}$$

Therefore,

$$\frac{dy}{dx} = \frac{1}{2\sqrt{x} + dy}$$

Because `$dy$`

is approaching zero, we can safely ignore it and we have

$$\frac{dy}{dx} = \frac{1}{2\sqrt{x}} = \frac{1}{2}x^{-\frac{1}{2}}$$

### Chain rule #

Let’s say we have `$f(x) = \sqrt{x}$`

, `$g(x) = x^2$`

, and `$h(x) = g(f(x)) = (\sqrt{x})^2 = x$`

.

We have:

\begin{align}
h^\prime(x) & = 1 = g^\prime(f(x)) \cdot f^\prime(x)\\

& = 2\cdot f(x) \cdot f^\prime(x) \\

& = 2\sqrt{x} \cdot f^\prime(x) \\

\end{align}

So we have

$$f^\prime(x) = \frac{1}{2\sqrt{x}}$$

The method by chain rule is inspired by Yifan Wei.

#MLLast modified on 2022-10-01