What is the derivative of sqrt x^2?

Jan 11, 2016

$\frac{x}{\sqrt{{x}^{2}}}$

Explanation:

first rewrite $\sqrt{{x}^{2}} = {\left({x}^{2}\right)}^{\frac{1}{2}}$

now differentiate using the chain rule :

$\frac{1}{2} {\left({x}^{2}\right)}^{- \frac{1}{2}} . \frac{d}{\mathrm{dx}} \left({x}^{2}\right)$

= 1/2(x^2)^(-1/2) .(2x) =x( x^2)^(-1/2) =x/(x^2)^(1/2) = x/sqrt(x^2

It is usual to give answer in the same form as question.

Jan 11, 2016

Assuming the question is asking ${\left(\sqrt{x}\right)}^{2}$ or $\sqrt{\left({x}^{2}\right)}$, which is then simplified to $x$, the answer is $1$.

Explanation:

${\left(\sqrt{x}\right)}^{2}$ can be rewritten as ${\left({x}^{\frac{1}{2}}\right)}^{2}$. Multiplying the exponents gives ${x}^{1}$, which is just $x$.

The definition of $\frac{d}{\mathrm{dx}} \left(x\right) = 1$ is typically memorized, though you can always use the power rule to get the same answer:

$\frac{d}{\mathrm{dx}} \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{1}\right) = \left(1\right) {x}^{1 - 1} = \left(1\right) \left({x}^{0}\right) = \left(1\right) \left(1\right) = 1$

If the problem were instead asking $\frac{d}{\mathrm{dx}} \left(\sqrt{x}\right)$, with the $2$ indicating that it's the second root, then the same power rule approach would be taken:

$\frac{d}{\mathrm{dx}} \left(\sqrt{x}\right) = \frac{d}{\mathrm{dx}} \left({x}^{\frac{1}{2}}\right) = \left(\frac{1}{2}\right) {x}^{\frac{1}{2} - 1} = \left(\frac{1}{2}\right) {x}^{- \frac{1}{2}} = \frac{1}{2 {x}^{\frac{1}{2}}}$
$= \frac{1}{2 \sqrt{x}}$