# What is the derivative of arctan(1+2x)^(1/2)?

Jul 13, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \left(1 + x\right) \sqrt{1 + 2 x}} .$

#### Explanation:

Let $y = \arctan {\left(1 + 2 x\right)}^{\frac{1}{2}}$

Put $u = {\left(1 + 2 x\right)}^{\frac{1}{2}}$ & $\left(1 + 2 x\right) = t .$

With these substitutions, we see that, y=arctanu, u=t^(1/2), &, t=1+2x...............(1).

Thus, $y$ is a function of $u$, $u$ of $t$, &, $t$ of $x$.

Accordingly, to find $\frac{\mathrm{dy}}{\mathrm{dx}}$ we need to apply the Chain Rule, which states that in the situation described above,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}$

={d/(du)(arctanu))}{d/dt(t^(1/2)}{d/dx(1+2x)}

$= \left\{\frac{1}{1 + {u}^{2}}\right\} \left\{\left(\frac{1}{2}\right) \cdot {t}^{- \frac{1}{2}}\right\} \left(0 + 2\right)$

$= \frac{1}{1 + {\left({\left(1 + 2 x\right)}^{\frac{1}{2}}\right)}^{2}} \left\{\frac{1}{t} ^ \left(\frac{1}{2}\right)\right\}$

$= \left\{\frac{1}{1 + 1 + 2 x}\right\} \cdot \left\{\frac{1}{1 + 2 x} ^ \left(\frac{1}{2}\right)\right\}$

$= \frac{1}{2 \left(1 + x\right)} \cdot \frac{1}{1 + 2 x} ^ \left(\frac{1}{2}\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \left(1 + x\right) \sqrt{1 + 2 x}} .$

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