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

Question

Samantha Adkison · Accepted Answer

#g'(x) = x(4x^2 -9x +4)#

The chain rule: #(dg(h(x)))/(dx)=(dg(h(x)))/(dh(x))*(dh(x))/(dx)#

To apply it to this problem, consider the other two terms as constants and calculate the derivative of one term at a time.

#g(x) = (x-1)(x-2)(x^2)#

#g'(x)= (d(x-1))/(dx)(x-2)(x^2) + (x-1)(d(x-2))/(dx)(x^2) + (x-1)(x-2)(d(x^2))/(dx)#

#g'(x) = (x-2)(x^2) + (x-1)(x^2) + (x-1)(x-2)(2x)#

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

#g'(x)=x(x^2-2x + x^2 - x + 2(x^2-3x+2))#

#g'(x) = x(4x^2 -9x +4)#

Charles Arocha · Answer

undefined To differentiate $ g(x) = (x-1)(x-2)(x^2) $ using the product rule, you first identify the functions $ f(x) $ and $ h(x) $ that are being multiplied together. Then you apply the product rule, which states that the derivative of the product of two functions is the derivative of the first function times the second function, plus the first function times the derivative of the second function.

Let $ f(x) = (x - 1)(x - 2) $ and $ h(x) = x^2 $.

Apply the product rule:

$$ g'(x) = f'(x)h(x) + f(x)h'(x) $$

First, find the derivatives of $ f(x) $ and $ h(x) $:

$$ f'(x) = (x - 1)'(x - 2) + (x - 1)(x - 2)' $$
$$ f'(x) = (1)(x - 2) + (x - 1)(1) $$
$$ f'(x) = x - 2 + x - 1 $$
$$ f'(x) = 2x - 3 $$

$$ h'(x) = (x^2)' $$
$$ h'(x) = 2x $$

Now, plug these derivatives back into the product rule formula:

$$ g'(x) = (2x - 3)(x^2) + (x - 1)(x - 2)(2x) $$

Simplify:

$$ g'(x) = 2x^3 - 3x^2 + 2x^3 - 6x^2 $$
$$ g'(x) = 4x^3 - 9x^2 $$

So, the derivative of $ g(x) = (x-1)(x-2)(x^2) $ using the product rule is $ g'(x) = 4x^3 - 9x^2 $.