Derivative of two polynomials, one of them being squared

[Orson]

Homework Statement

find derivative of (x-2)(x-3)^2

Homework Equations

using product rule.

The Attempt at a Solution

1(x-3)^2+2(x-3)
x^2-6x-9 +2x-6
x^2-4x-15
doesn't factor.

 

[fresh_42]

Homework Statement

find derivative of (x-2)(x-3)^2

Homework Equations

using product rule.

The Attempt at a Solution

1(x-3)^2+2(x-3)
x^2-6x-9 +2x-6
x^2-4x-15
doesn't factor.

There are a few mistakes in there. You've forgotten a factor in the product rule $(g\cdot h)'=g' \cdot h + g \cdot h'$ and a sign error in the binomial formula.

 

[Orson]

Did I forget the 1?

 

[Orson]

tried again and 3x^2-14x+21

 

[Orson]

for clarification: using product rule.

(x-2)(x-3)^2
1(x-3)+(x-2)2(x-3)
(x-3)+(x-2)2x-6)
(x-3)+2x^2-10x+12
2x^2-9x+9
(2x-3) (x-3)
x=3
x=3/2

 

[fresh_42]

Did I forget the 1?

No, the $g$.

tried again and 3x^2-14x+21

Close.

You can always calculate $p(x)=(x-2)(x-3)^2$ and differentiate the result to have a comparison.

 

[fresh_42]

for clarification: using product rule.

(x-2)(x-3)^2
1(x-3)+(x-2)2(x-3)
(x-3)+(x-2)2x-6)
(x-3)+2x^2-10x+12
2x^2-9x+9
(2x-3) (x-3)
x=3
x=3/2

This time the square in $h$ is lost, i.e. the second line should be $1\cdot (x-3)^2 + (x-2)\cdot 2 \cdot (x-3) \cdot 1$.

 

[Orson]

This time the square in $h$ is lost, i.e. the second line should be $1\cdot (x-3)^2 + (x-2)\cdot 2 \cdot (x-3) \cdot 1$.

you're right. I will chew on it some more.

 

[epenguin]

It has got to factor and the factor has got to be (x - 3).

Instead of getting lost in error-prone detail, differentiate the more general fg²

Does the result of that factorise? Easily?

 

[Orson]

It has got to factor and the factor has got to be (x - 3).

Instead of getting lost in error-prone detail, differentiate the more general fg²

Does the result of that factorise? Easily?

certainly does factor.

(x-3)^2 +(x-2)*2(x-3)
x^2-6x+9+(x-2)(2x-6)
x^2-6x+9+2x^2-10x+12
3x^2-16x+21
3x^2-9x-7x+21
3x(x-3)-7(x-3)
(3x-7)(x-3)
x=7/3 x=3

 

[fresh_42]

That's correct.

 

[WWGD]

certainly does factor.

(x-3)^2 +(x-2)*2(x-3)
x^2-6x+9+(x-2)(2x-6)
x^2-6x+9+2x^2-10x+12
3x^2-16x+21
3x^2-9x-7x+21
3x(x-3)-7(x-3)
(3x-7)(x-3)
x=7/3 x=3

I don't see why you're finding the zeros of the expression.

 

[Mark44]

I don't see why you're finding the zeros of the expression.

I don't either, based on the problem description at the start of the thread. However, if another part of the problem is to find the critical points, then one would need to factor the derivative to see where it is zero.

 

[Orson]

I don't see why you're finding the zeros of the expression.

Yes. I was finding maxima and minima. I had trouble factoring that function.

 

[epenguin]

But there is no virtue in doing it a long way round when you can do it in a way that retains an original factor.

 

[Orson]

But there is no virtue in doing it a long way round when you can do it in a way that retains an original factor.

Ok. Can you expound on that ?

 

[Mark44]

Ok. Can you expound on that ?

I think what @epenguin was saying was that it's not helpful to expand the factors that you get when you differentiate. If y = (x - 2)(x - 3)²,
then y' = (x - 2) * 2(x - 3) + 1 * (x - 3)², which can by rewritten as (x - 3)(2(x - 2) + x - 3)) = (x - 3)(3x - 7)
Now to find where y' = 0, it's easy to see that x = 3 or x = 7/3.

 

[epenguin]

Yes IMO still easier to see if you keep the more general form fg².
Just use differentiation of a product rule (uv)' = u'v + uv' which applied to fg² is

( fg²)' = f'g² + 2fgg' = g(f' + 2fg') .

Something like this is always going to happen if the original function contains something to a power.

 

[Orson]

Yes IMO still easier to see if you keep the more general form fg².
Just use differentiation of a product rule (uv)' = u'v + uv' which applied to fg² is

( fg²)' = f'g² + 2fgg' = g(f' + 2fg') .

Something like this is always going to happen if the original function contains something to a power.

Lemme chew on that. I have another one I am having trouble with. Has to do with minus signs.