# What is the derivative of sqrt(x^2+2x-1)?

Mar 5, 2016

$\frac{x + 1}{\sqrt{{x}^{2} + 2 x - 1}}$

#### Explanation:

What we have here is a function within a function; ${x}^{2} + 2 x - 1$ is under the radical ($\sqrt{}$) sign. That means we have to use the chain rule to differentiate, which says that you take the derivative of the "inside" function (in this case ${x}^{2} + 2 x - 1$) and multiply it by the derivative of the whole function.

Begin by finding the derivative of ${x}^{2} + 2 x - 1$. Using the power rule, the derivative is $2 x + 2$. Now onto the whole function. Note that we can write $\sqrt{{x}^{2} + 2 x - 1}$ as ${\left({x}^{2} + 2 x - 1\right)}^{\frac{1}{2}}$. That means we can again apply the power rule:
$\frac{d}{\mathrm{dx}} {\left({x}^{2} + 2 x - 1\right)}^{\frac{1}{2}} = \frac{1}{2 {\left({x}^{2} + 2 x - 1\right)}^{\frac{1}{2}}}$

Now we can multiply this by the derivative of the inside function, which we found as $2 x + 2$. Performing this operation yields:
$\frac{1}{2 {\left({x}^{2} + 2 x - 1\right)}^{\frac{1}{2}}} \cdot 2 x + 2 = \frac{2 x + 2}{2 {\left({x}^{2} + 2 x - 1\right)}^{\frac{1}{2}}}$

Finally, look for any ways to simplify the problem. We see that there is a $2$ in the denominator - is there any way we can get rid of it? In fact, there is by factoring out a $2$ from the numerator; take a look:
$\frac{2 \left(x + 1\right)}{2 {\left({x}^{2} + 2 x - 1\right)}^{\frac{1}{2}}} = \frac{x + 1}{{\left({x}^{2} + 2 x - 1\right)}^{\frac{1}{2}}}$

Because the problem was given to us in radical form, we should convert it back, rewriting the answer as:
$\frac{x + 1}{\sqrt{{x}^{2} + 2 x - 1}}$