## Proof

$$\sqrt{x}={x}^{\frac{1}{2}}$$**Definition of $\sqrt{x}$:**

$$\text{If}y={x}^{n},\text{then}{y}^{\prime}=n{x}^{n-1}$$**Using the power rule:**Let $n=\frac{1}{2}$:

$$\frac{\text{d}}{\text{d}x}{x}^{\frac{1}{2}}=\frac{1}{2}{x}^{\frac{1}{2}-1}$$

$$\frac{\text{d}}{\text{d}x}{x}^{\frac{1}{2}}=\frac{1}{2}{x}^{-\frac{1}{2}}$$**Simplify the expression:**

$${x}^{-\frac{1}{2}}=\frac{1}{\sqrt{x}}$$**Rewrite using the square root:**

Thus, the derivative of $\sqrt{x}$ is:

$$\frac{\text{d}}{\text{d}x}\sqrt{x}=\overline{)\frac{1}{2\sqrt{x}}}$$## Explanation

To understand the derivative of $\sqrt{x}$, we start by expressing $\sqrt{x}$ as a power of $x$. Specifically, $\sqrt{x}$ can be written as ${x}^{\frac{1}{2}}$. This form allows us to use the power rule for differentiation.

The power rule states that if we have a function ${x}^{n}$, its derivative with respect to $x$ is $n{x}^{n-1}$. Here, our exponent $n$ is $\frac{1}{2}$.

Applying the power rule to ${x}^{\frac{1}{2}}$, we get:

$$\frac{\text{d}}{\text{d}x}{x}^{\frac{1}{2}}=\frac{1}{2}{x}^{\frac{1}{2}-1}$$Next, we simplify the exponent $\frac{1}{2}-1$, which equals $-\frac{1}{2}$. Therefore, our expression for the derivative becomes:

$$\frac{1}{2}{x}^{-\frac{1}{2}}$$To make this expression more familiar, we rewrite ${x}^{-\frac{1}{2}}$ using the square root notation. Since ${x}^{-\frac{1}{2}}$ is the same as $\frac{1}{{x}^{\frac{1}{2}}}$, and ${x}^{\frac{1}{2}}$ is $\sqrt{x}$, we get:

$${x}^{-\frac{1}{2}}=\frac{1}{\sqrt{x}}$$Substituting this back into our expression for the derivative, we have:

$$\frac{\text{d}}{\text{d}x}{x}^{\frac{1}{2}}=\frac{1}{2}\xb7\frac{1}{\sqrt{x}}=\frac{1}{2\sqrt{x}}$$Therefore, the derivative of $\sqrt{x}$ with respect to $x$ is $\overline{)\frac{1}{2\sqrt{x}}}$.

*Q.E.D.*