What is the derivative of #f(x)=(x-1)(x-2)(x-3)#?

Question

Luke Alvarez · Accepted Answer

#f'(x)=3x^2-12x+11#.

I'll go over two approaches to solving the issue. Approach I: We use the Multiplication Rule for Differentiation # : #(uvw)'=u'vw+v'uw+uvw'# Hence, #f'(x)=(x-1)'(x-2)(x-3)+(x-1)(x-2)'(x-3)+(x-1)(x-2)(x-3)'# #=(x-2)(x-3)+(x-1)(x-3)+(x-1)(x-2)# #f'(x)=3x^2-12x+11#. Technique II: Given, #f(x)=(x-1)(x-2)(x-3)=x^3-6x^2+11x-6#. Hence, #f"(x)=3x^2-12x+11#, as Before! Have fun with math!

Cameron Alexander · Answer

undefined To find the derivative of $ f(x) = (x-1)(x-2)(x-3) $, you can use the product rule. The product rule states that if $ u(x) $ and $ v(x) $ are differentiable functions of $ x $, then the derivative of their product is given by $ (u \cdot v)' = u'v + uv' $. Applying the product rule to $ f(x) $, we have:

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

Now, find the derivatives of $ (x-2) $ and $ (x-3) $ using the power rule, which states that $ \frac{d}{dx}(x^n) = nx^{n-1} $:

$ (x-2)' = 1 \cdot (x-2)^{1-1} = 1 $

$ (x-3)' = 1 \cdot (x-3)^{1-1} = 1 $

Substitute these derivatives back into the expression:

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

Simplify:

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

Now, expand the terms:

$ f'(x) = x^2 - 4x + 3 + x^2 - 3x + 2 $

Combine like terms:

$ f'(x) = 2x^2 - 7x + 5 $

So, the derivative of $ f(x) = (x-1)(x-2)(x-3) $ is $ f'(x) = 2x^2 - 7x + 5 $.