At first, it may appear daunting to calculate the derivative of $\sqrt x$. Indeed, you have not seen anything like this before! You know the power rule, the derivatives of trigonometric functions, like derivative of sine and derivative of cosine. You might even remember the derivatives of tangent, cotangent, secant and cosecant. Unexpectedly, even the derivatives of inverse trigonometric functions may come to mind. However, $\sqrt x$ does not appear to be any of the derivatives you know. So what to do? Surprisingly, laws of exponents come to the rescue. We can write

$$\sqrt x = x^{\frac12}$$

and now we can use the power rule:

$$\dfrac d{dx}(\sqrt x) = \dfrac d{dx}(x^{\frac12}) = \frac12x^{-\frac12} = \frac12\frac1{x^{\frac12}} = \frac1{2\sqrt x}$$

Thus, we conclude that $\boxed{\dfrac{d}{dx}(\sqrt x) = \dfrac1{2\sqrt x}}$.

### Laws of Exponents and Radicals

Using laws of exponents is a powerful technique when it comes to differentiating expressions involving radicals. For example, if we wish to differentiate $x^2\sqrt x$, it might appear that we should use the product rule, but it would be so much simpler to rewrite the expression using the laws of exponents and then use the power rule:

$$x^2\sqrt x = x^2 x^{\frac12} = x^{2 + \frac12} = x^{\frac52}$$

For comparison, the power rule yields the answer immediately

$$\frac d{dx}(x^{\frac52}) = \frac52 x^{\frac32}$$

On the other hand, if we use the product rule, namely,

$$\dfrac d{dx} (f(x)g(x)) = f'(x) g(x) + f(x) g'(x)$$

with $f(x) = x^2$ and $g(x) = \sqrt x$ and $f'(x) = 2x$ and $g'(x) = \dfrac1{2\sqrt x}$, we get

$$2x\sqrt x + x^2\dfrac1{2\sqrt x}$$

Not only is this expression more complicated, it is not immediately clear how to combine it. The conclusion is that you should always try to use the power rule when differentiating (and integrating!) expressions that involve powers of $x$. Powers of $x$ may involve radicals, so you should always remember to use the laws of exponents.