# How do you differentiate g(x) = sqrtxsqrt(1-e^(2x)) using the product rule?

Nov 23, 2017

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

#### Explanation:

The product rule states that:
$\frac{d}{\mathrm{dx}} \left(f \left(x\right) g \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) g \left(x\right) + f \left(x\right) \frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)$

Plugging in your above expression, we get:
$\frac{d}{\mathrm{dx}} \left(\sqrt{x} \sqrt{1 - {e}^{2 x}}\right) = \frac{d}{\mathrm{dx}} \left(\sqrt{x}\right) \sqrt{1 - {e}^{2 x}} + \sqrt{x} \frac{d}{\mathrm{dx}} \left(\sqrt{1 - {e}^{2 x}}\right)$

The $\sqrt{x}$ derivative is quite easy if you notice that $\sqrt{x} = {x}^{\frac{1}{2}}$ and use the power rule. However, the other derivative is a bit more complex. We need to use the chain rule to evaluate it:
$\frac{d}{\mathrm{dx}} \left(\sqrt{1 - {e}^{2 x}}\right)$

We can let $u = 1 - {e}^{2 x}$, and rewrite the expression like this:
$\frac{d}{\mathrm{du}} \left(\sqrt{u}\right) \frac{d}{\mathrm{dx}} \left(1 - {e}^{2 x}\right) = \frac{1}{2 \sqrt{u}} \cdot \frac{d}{\mathrm{dx}} \left(1 - {e}^{2 x}\right)$

Now we once again need to use the chain rule. If we let $z = 2 x$, we get:
$\frac{1}{2 \sqrt{u}} \cdot \frac{d}{\mathrm{dz}} \left(1 - {e}^{z}\right) \frac{d}{\mathrm{dx}} \left(2 x\right) = - \frac{1}{2 \sqrt{u}} 2 {e}^{z}$

Resubstituting, we get:
$- \frac{1}{2 \sqrt{1 - {e}^{2 x}}} \cdot 2 {e}^{2 x} = - {e}^{2 x} / \sqrt{1 - {e}^{2 x}}$

Now that we have evaluated the difficult part in our product rule expression, we can put it back in and see that the answer becomes this:
$\frac{1}{2 \sqrt{x}} \sqrt{1 - {e}^{2 x}} - \sqrt{x} {e}^{2 x} / \sqrt{1 - {e}^{2 x}} = \frac{\sqrt{1 - {e}^{2 x}}}{2 \sqrt{x}} - \frac{\sqrt{x} {e}^{2 x}}{\sqrt{1 - {e}^{2 x}}}$

If we multiply by the opposite denominator, we can get the two fractions to have the same denominator:
$\frac{\sqrt{1 - {e}^{2 x}} \sqrt{1 - {e}^{2 x}}}{2 \sqrt{x} \sqrt{1 - {e}^{2 x}}} - \frac{2 \sqrt{x} \sqrt{x} {e}^{2 x}}{2 \sqrt{x} \sqrt{1 - {e}^{2 x}}} = \frac{1 - {e}^{2 x} - 2 x {e}^{2 x}}{2 \sqrt{x} \sqrt{1 - {e}^{2 x}}}$