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

Question

Aurora Alvarado · Accepted Answer

#y^' = (2x)/(1-x^2)^2#

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

#1/(1-x^2) = (1-x^2)^(-1) = 1 * (1-x^2)^(-1)#

You can now use the formula

#color(blue)(d/dx(y) = [d/dx(f(x))] * g(x) + f(x) * d/dx(g(x)))#

In your case, you have #f(x) = 1# a nd #g(x) = (1-x^2)^(-1)#.

This means that the derivative of #y# will be

#d/dx(y) = [d/dx(1)] * (1-x^2)^(-1) + 1 * d/dx(1-x^2)^(-1)#

Now, to differentiate #(1-x^2)^(-1)# you can use the chain rule for #u^(-1)#, with #u = 1-x^2#. This will get you

#d/dx(u^(-1)) = d/(du)u^(-1) * d/dx(u)#

#d/dx(u^(-1)) = -u^(-2) * d/dx(1-x^2)#

#d/dx(u^(-1)) = -u^(-2) * (-2x)#

#d/dx(1-x^2)^(-1) = -(1-x^2)^(-2) * (-2x)#

#d/dx(1-x^2)^(-1) = 2x * (1-x^2)^(-2)#

Your target derivative will thus be

#y^' = 0 * (1-x^2)^(-1) + 1 * 2x * (1-x^2)^(-2)#

#y^' = color(green)((2x)/(1-x^2)^2)#

Cameron Alexander · Answer

undefined To use the Product Rule to find the derivative of $ y = \frac{1}{1-x^2} $, follow these steps:

1. Identify the two functions being multiplied: $ f(x) = 1 $ and $ g(x) = \frac{1}{1-x^2} $.
2. Apply the Product Rule, which states that if $ y = f(x) \cdot g(x) $, then $ y' = f'(x) \cdot g(x) + f(x) \cdot g'(x) $.
3. Find the derivatives of $ f(x) $ and $ g(x) $ using the power rule and chain rule, respectively.
   - $ f'(x) = 0 $ (since the derivative of a constant is zero).
   - $ g'(x) = \frac{d}{dx}(\frac{1}{1-x^2}) $.
4. Use the chain rule to find $ g'(x) $.
   - Let $ u = 1-x^2 $.
   - $ \frac{du}{dx} = -2x $.
   - $ g'(x) = \frac{d}{du}(\frac{1}{u}) \cdot \frac{du}{dx} $.
   - Substitute $ u = 1-x^2 $ and $ \frac{du}{dx} = -2x $ into $ g'(x) $.
5. Plug the derivatives of $ f(x) $ and $ g(x) $ into the Product Rule formula.
6. Simplify the expression to get the final derivative.