How do you differentiate #f(x) = (x-1)(x-2)# using the product rule?

Question

Emma Aldridge · Accepted Answer

The product rule states that for a function #f(x) = g(x)h(x)#, #f'(x) = (g'(x) xx h(x)) + (h'(x) xx g(x))#

In our case, let #f(x) = g(x) xx h(x)#

Therefore, #g(x) = x - 1# and #h(x) = x - 2#

Let's examine each of these functions separately.

According to the power rule:

#g'(x) = 1#

and

#h'(x) = 1#

We are now able to use the product rule.

#f'(x) = (g'(x) xx h(x)) + (h'(x) xx g(x))#

#f'(x) = ((x - 2) xx 1) + ((x - 1) xx 1)#

#f'(x) = x - 2 + x - 1#

#f'(x) = 2x - 3#

Therefore #dy/dx = 2x - 3#

I hope this is useful!

Charles Anctil · Answer

undefined To differentiate $ f(x) = (x-1)(x-2) $ using the product rule:

1. Identify the two functions being multiplied: $ f(x) = (x-1) $ and $ g(x) = (x-2) $.
2. Apply the product rule formula: $ (f \cdot g)' = f' \cdot g + f \cdot g' $.
3. Differentiate each function individually:
   - $ f'(x) = 1 $ (differentiation of $ x-1 $)
   - $ g'(x) = 1 $ (differentiation of $ x-2 $).
4. Substitute these differentials into the product rule formula:
   $ (f \cdot g)' = (1)(x-2) + (x-1)(1) $.
5. Simplify the expression:
   $ (f \cdot g)' = x - 2 + x - 1 $.
6. Combine like terms:
   $ (f \cdot g)' = 2x - 3 $.

Therefore, the derivative of $ f(x) = (x-1)(x-2) $ using the product rule is $ f'(x) = 2x - 3 $.