# How do you use the Product Rule to find the derivative of  y = 1/(1-x^2)?

Aug 9, 2015

${y}^{'} = \frac{2 x}{1 - {x}^{2}} ^ 2$

#### Explanation:

In order to differentiate this function by using the product rule, you need to find a way to write it as a product of two functions.

Notice that you can write the function as

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

You can now use the formula

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(y\right) = \left[\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right)\right] \cdot g \left(x\right) + f \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)}$

In your case, you have $f \left(x\right) = 1$ a nd $g \left(x\right) = {\left(1 - {x}^{2}\right)}^{- 1}$.

This means that the derivative of $y$ will be

$\frac{d}{\mathrm{dx}} \left(y\right) = \left[\frac{d}{\mathrm{dx}} \left(1\right)\right] \cdot {\left(1 - {x}^{2}\right)}^{- 1} + 1 \cdot \frac{d}{\mathrm{dx}} {\left(1 - {x}^{2}\right)}^{- 1}$

Now, to differentiate ${\left(1 - {x}^{2}\right)}^{- 1}$ you can use the chain rule for ${u}^{- 1}$, with $u = 1 - {x}^{2}$. This will get you

$\frac{d}{\mathrm{dx}} \left({u}^{- 1}\right) = \frac{d}{\mathrm{du}} {u}^{- 1} \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

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

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

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

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

Your target derivative will thus be

${y}^{'} = 0 \cdot {\left(1 - {x}^{2}\right)}^{- 1} + 1 \cdot 2 x \cdot {\left(1 - {x}^{2}\right)}^{- 2}$

${y}^{'} = \textcolor{g r e e n}{\frac{2 x}{1 - {x}^{2}} ^ 2}$