Algebra simplification after derivation problem

[Andrew Rubin]

I'm working on a derivation problem where I get the right answer, but I struggle in further simplifying my answer,

[MATH]f\left(x\right)\:=\:\left(x^2-4x+3\right)e^{\frac{1}{2}x}[/MATH]
I get,

[MATH]f'\left(x\right)=\left(2x-4\right)\cdot \:e^{\frac{1}{2}x}+\left(x^2-4x+3\right)\cdot \frac{1}{2}\cdot \:e^{\frac{1}{2}x}[/MATH]
Which according to my textbook's solution is correct. However, it can be further simplified - and this is where I fall off:

[MATH]=\:\left(\frac{1}{2}x^2-\frac{5}{2}\right)e^{\frac{1}{2}x}[/MATH]
[MATH]=\frac{1}{2}\left(x^2-5\right)e^{\frac{1}{2}x}[/MATH]
I have some trouble understanding which algebra rules to apply in the last two steps as well as the intuition. I would greatly appreciate input as well as links that may give some overview of what algebra area I'm not seeing here. My guess is it's pretty basic, but it still requires some refreshing for my part.

 

[Deleted member 4993]

I'm working on a derivation problem where I get the right answer, but I struggle in further simplifying my answer,

[MATH]f\left(x\right)\:=\:\left(x^2-4x+3\right)e^{\frac{1}{2}x}[/MATH]
I get,

[MATH]f'\left(x\right)=\left(2x-4\right)\cdot \:e^{\frac{1}{2}x}+\left(x^2-4x+3\right)\cdot \frac{1}{2}\cdot \:e^{\frac{1}{2}x}[/MATH]
Which according to my textbook's solution is correct. However, it can be further simplified - and this is where I fall off:

[MATH]=\:\left(\frac{1}{2}x^2-\frac{5}{2}\right)e^{\frac{1}{2}x}[/MATH]
[MATH]=\frac{1}{2}\left(x^2-5\right)e^{\frac{1}{2}x}[/MATH]
I have some trouble understanding which algebra rules to apply in the last two steps as well as the intuition. I would greatly appreciate input as well as links that may give some overview of what algebra area I'm not seeing here. My guess is it's pretty basic, but it still requires some refreshing for my part.

Do you see:

\(\displaystyle (2x - 4)*e^{\frac{x}{2}} + \frac{1}{2} * (x^2 - 4x +3) * e^{\frac{x}{2}}\)

=\(\displaystyle (2x - 4)*e^{\frac{x}{2}} +( \frac{1}{2} * x^2 - \frac{1}{2} * 4x + \frac{1}{2} * 3) * e^{\frac{x}{2}}\)

=\(\displaystyle (2x - 4)*e^{\frac{x}{2}} +( \frac{x^2}{2} - 2x + \frac{3}{2}) * e^{\frac{x}{2}}\)

Now factorize e^(x/2) and continue.

 

[JeffM]

What you are missing is the ability to "see" potential application of the law of distribution of multiplication over addition in complex cases.

[MATH]a(b \pm c) \equiv ab \pm ac.[/MATH]
I greatly doubt that you are unaware of the law, which is exemplied by

[MATH]7(3 + 8) = 21 + 56 = 77 \text { and } 7(3 + 8) = 7 * 11 = 77.[/MATH]
But you do not look for common factors in more comlicated expressions.

 

[Andrew Rubin]

This was great feedback. When combining your answers I see that my problem lies in applying a rule I'm familiar with in a relatively new context for me. I also follow your steps @Subhotosh Khan, however, it would be really great if you could show how the remaining steps results in the "simplified" solutions in my first post. I just want to make sure that I understand the remaining steps as well.

 

[JeffM]

[MATH](2x - 4) * e^{x/2} + \left ( \dfrac{x^2}{2} - 2x + \dfrac{3}{2} \right )* e^{x/2} = [/MATH]
[MATH]e^{x/2} \left \{ (2x - 4) + \left ( \dfrac{x^2}{2} - 2x + \dfrac{3}{2} \right ) \right \} = [/MATH]
[MATH]e^{x/2} \left \{ 1 * (2x - 4) + \left ( \dfrac{1}{2} * x^2 - (1)(2x) + \dfrac{1}{2} * 3 \right ) \right \} = [/MATH]
[MATH]e^{x/2} \left \{ \dfrac{2}{2} * (2x - 4) + \left ( \dfrac{1}{2} * x^ 2 - \dfrac{2}{2} * 2x + \dfrac{1}{2} * 3 \right ) \right \} =[/MATH]
[MATH]e^{x/2} \left \{ \dfrac{1}{2} * (4x - 8) + \left ( \dfrac{1}{2} * x^2 - \dfrac{1}{2} * 4x + \dfrac{1}{2} * 3 \right ) \right \} =[/MATH]
[MATH]e^{x/2} \left \{ \dfrac{1}{2} * (4x - 8) + \dfrac {1}{2} * (x^2 - 4x + 3 ) \right \}= [/MATH]
[MATH]\dfrac{1}{2} * e^{x/2} * ( 4x - 8 + x^2 - 4x + 3) = [/MATH]
[MATH]\dfrac{1}{2} * e^{x/2} * (x^2 - 5) = \dfrac{\sqrt{e^x} * (x^2 - 5)}{2}.[/MATH]

 

[Deleted member 4993]

This was great feedback. When combining your answers I see that my problem lies in applying a rule I'm familiar with in a relatively new context for me. I also follow your steps @Subhotosh Khan, however, it would be really great if you could show how the remaining steps results in the "simplified" solutions in my first post. I just want to make sure that I understand the remaining steps as well.

The first step of the remaining steps is to - factorize e^(-x/2) - the common factor - out.

Did you do that?

What "expression did you get?

 

[Andrew Rubin]

Thanks again for your replies - this is incredible valuable for me as I've struggled to find good explanations elsewhere! I really appreciate you using your time to answer these questions.

@Subhotosh Khan, the expression I got was:

[MATH]e^{\frac{1}{2}x}\left(\left(2x−4\right)+\left(\frac{x^2}{2}−2x+\frac{3}{2}\right)\right)[/MATH]
I get every step except two, @JeffM. If you would kindly explain these two, I think I've finally cracked this:

First, in

[MATH]e^{\frac{1}{2}x}\left(1\cdot \left(2x−4\right)+\left(\frac{1}{2}\cdot x^2−\left(1\right)\left(2x\right)+\frac{1}{2}\cdot 3\right)\right)[/MATH]
How do you arrive at:
[MATH]1\cdot(2x−4)[/MATH], [MATH]\frac{1}{2}\cdot x^2−\left(1\right)[/MATH], and [MATH]\frac{1}{2}\cdot 3[/MATH]. I don't fully understand where 1 and 3 is coming from here.

Second, in going from

[MATH]e^{\frac{1}{2}x}\left(\frac{2}{2}\cdot \left(2x−4\right)+\left(\frac{1}{2}\cdot x^2-\frac{2}{2}\cdot 2x+\frac{1}{2}\cdot 3\right)\right)[/MATH],

to,

[MATH]e^{\frac{1}{2}x}\left(\frac{1}{2}\cdot \left(4x−8\right)+\left(\frac{1}{2}\cdot x^2-\frac{1}{2}\cdot 4x+\frac{1}{2}\cdot 3\right)\right)[/MATH]
I see that the expressions are multiplied by 2 using [MATH]\frac{2}{2}[/MATH], but I can't quite grasp why [MATH]\frac{1}{2}[/MATH] is still remaining in the last expression.

 

[JeffM]

[USER]@JeffM[/USER]. If you would kindly explain these two, I think I've finally cracked this:

First, in

[MATH]e^{\frac{1}{2}x}\left(1\cdot \left(2x−4\right)+\left(\frac{1}{2}\cdot x^2−\left(1\right)\left(2x\right)+\frac{1}{2}\cdot 3\right)\right)[/MATH]
How do you arrive at:
[MATH]1\cdot(2x−4)[/MATH], [MATH]\frac{1}{2}\cdot x^2−\left(1\right)[/MATH], and [MATH]\frac{1}{2}\cdot 3[/MATH]. I don't fully understand where 1 and 3 is coming from here.

Second, in going from

[MATH]e^{\frac{1}{2}x}\left(\frac{2}{2}\cdot \left(2x−4\right)+\left(\frac{1}{2}\cdot x^2-\frac{2}{2}\cdot 2x+\frac{1}{2}\cdot 3\right)\right)[/MATH],

to,

[MATH]e^{\frac{1}{2}x}\left(\frac{1}{2}\cdot \left(4x−8\right)+\left(\frac{1}{2}\cdot x^2-\frac{1}{2}\cdot 4x+\frac{1}{2}\cdot 3\right)\right)[/MATH]
I see that the expressions are multiplied by 2 using [MATH]\frac{2}{2}[/MATH], but I can't quite grasp why [MATH]\frac{1}{2}[/MATH] is still remaining in the last expression.

When you first start learning algebra, solving equations is largely a matter of simplifying and re-arranging terms. In trigonometry and calculus, it is often neccesary to transform expressions into apparently more complex forms that are in fact easier to work with. Two common ways that is done are

[MATH]x = 1 * x = \dfrac{a}{a} * x = \dfrac{ax}{a} \text { provided } a \ne 0, [/MATH] and

[MATH]x = x + 0 = x + (c - c) = (x + c) - c.[/MATH]
These are general techniques used, for example, in completing the square. Now to get specific. I hate working with fractions unless I have to. And I can "see" that

[MATH]\dfrac{x^2}{2} = \dfrac{x^2}{1} * \dfrac{1}{2} = x^2 * \dfrac{1}{2} \text { and } \dfrac{3}{2} = \dfrac{3}{1} * \dfrac{1}{2} = 3 * \dfrac{1}{2}.[/MATH]
Two of my terms contain a factor of 1/2 by just an iota of re-arrangement. So now you should see where the 3 is coming from. If all my terms contained a factor of 1/2, I could factor 1/2 out and simplify by eliminating all fractions inside the parentheses. So I use the technique of multiplying by a useful form of 1, in this case 2/2.

[MATH]u = 1 * u = \dfrac{2}{2} * u = \dfrac{2}{2} * \dfrac{u}{1} = \dfrac{1}{2} * \dfrac{2u}{1} = \dfrac{1}{2} * 2u.[/MATH]
So, for example,

[MATH]2x - 4 = 1 * (2x - 4) = \dfrac{2}{2} * (2x - 4) = \dfrac{2}{2} * \dfrac{2x - 4}{1} = \dfrac{1}{2} * \dfrac{2(2x - 4)}{1} = \dfrac{1}{2} * \dfrac{4x - 8}{1} = \dfrac{1}{2} * (4x - 8).[/MATH]
The expressions are not multiplied by 2, but by 1, which does not change their value. It is just that 1 is in the form of

[MATH]1 = \dfrac{2}{2} = \dfrac{1}{2} * 2.[/MATH]
It is a simple technique but a helpful one. You have seen it before in rationalizing fractions.

[MATH]\dfrac{2}{\sqrt{5} - \sqrt{3}} = \dfrac{2}{\sqrt{5} - \sqrt{3}} * 1 = \dfrac{2}{\sqrt{5} - \sqrt{3}} * \dfrac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}} =[/MATH]
[MATH]\dfrac{2(\sqrt{5} + \sqrt{3})}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})} = \dfrac{2(\sqrt{5} + \sqrt{3})}{(\sqrt{5})^2 - (\sqrt{3})^2} = \dfrac{2(\sqrt{5} + \sqrt{3})}{5 - 3} =[/MATH]
[MATH]\dfrac{2(\sqrt{5} + \sqrt{3})}{2} = \sqrt{5} + \sqrt{3}.[/MATH]
A personal note: I got good at algebra by studying calculus.

 

[Andrew Rubin]

Thank you so much @JeffM! Now I'm able to understand every step, making it easier to continue my work. Math is really fun when the pieces start to click.