Shorten equation in derivation problem

[Andrew Rubin]

Hi everyone

I'm currently working through an introductory calculus textbook, where I'm getting the right answers, but I'm stuck at shortening the answers. I suspect I have to work more on my algebra, but I would be very happy for some pointers on what I'm missing. The common problem seem to be shortening with brackets involved. Particulary, for the chapter I'm working through, I have two examples:

1. Find the derivate of [MATH]f\left(x\right)=\left(x^2+1\right)e^x[/MATH]
Where I get the answer [MATH]f'\left(x\right)=2xe^2+\left(x^2+1\right)e^x[/MATH]
which is correct according to the book, but can be shortened to,

[MATH]f'\left(x\right)=\left(x^2+2x+1\right)e^x[/MATH]
2. Find the derivate of [MATH]f\left(x\right)\:=\:\left(2x-1\right)e^x[/MATH]
I get,

[MATH]f'\left(x\right)=2e^x\:+\:\left(2x-1\right)e^x[/MATH]
that can be shortened to,

[MATH]f'\left(x\right)\:=\:\left(2x+1\right)e^x[/MATH]

 

[tkhunny]

There is more going on that you are suggesting. Your first clue should be your result contain [math]e^{2}[/math]. That's no good.

If [math]f(x) = g(x)\cdot h(x)\;then\;f'(x) = g(x)\cdot h'(x)+g'(x)\cdot h(x)[/math]
You have [math]g(x) = x^{2} + 1\;and\;h(x) = e^{x}[/math].

Carefully find the two derivatives and carefully assemble the formulaic result.

The "shortened" versions are from algebra alone. This is not a calculus concept. Review?

 

[JeffM]

Hi everyone

I'm currently working through an introductory calculus textbook, where I'm getting the right answers, but I'm stuck at shortening the answers. I suspect I have to work more on my algebra, but I would be very happy for some pointers on what I'm missing. The common problem seem to be shortening with brackets involved. Particulary, for the chapter I'm working through, I have two examples:

1. Find the derivate of [MATH]f\left(x\right)=\left(x^2+1\right)e^x[/MATH]
Where I get the answer [MATH]f'\left(x\right)=2xe^2+\left(x^2+1\right)e^x[/MATH]
which is correct according to the book, but can be shortened to,

[MATH]f'\left(x\right)=\left(x^2+2x+1\right)e^x[/MATH]
2. Find the derivate of [MATH]f\left(x\right)\:=\:\left(2x-1\right)e^x[/MATH]
I get,

[MATH]f'\left(x\right)=2e^x\:+\:\left(2x-1\right)e^x[/MATH]
that can be shortened to,

[MATH]f'\left(x\right)\:=\:\left(2x+1\right)e^x[/MATH]

First, you made an error in your first derivative.

[MATH]f(x) = (x^2 + 1)e^x \implies f'(x) = 2xe^x + (x^2 + 1)e^x \ne 2xe^2 + (x^2 + 1).[/MATH]
You say that you got the correct answer so I am hoping that what you wrote in your post was a typo. Using the correct answer, do you see that [MATH]e^x[/MATH] is a common factor of both summands?

[MATH] 2xe^x + (x^2 + 1)e^x = e^x * 2x + e^x * (x^2 + 1) = e^x(2x + x^2 + 1) = e^x(x^2 + 2x + 1) = e^x(x + 1)^2.[/MATH]
The same issue applies to the second problem.

This is basic simplification in algebra.

 

[HallsofIvy]

I would assume that the "\(\displaystyle e^2\)" instead of "\(\displaystyle e^x\)" is a typo. The simplifications that you are asking about are just a matter of factoring out \(\displaystyle e^x\).

\(\displaystyle 2xe^x+ (x^2+ 1)e^x= (2x+ x^2+ 1)e^x= (x^2+ 2x+ 1)e^x= (x+ 1)^2e^x\).

\(\displaystyle 2e^x+ (2x- 1)e^x= (2+ 2x- 1)e^x= (2x+ 1)e^x\).

 

[Andrew Rubin]

Thanks for all your helpful replies! I see that this clearly falls under algebra, not derivation, so I'll try to edit it. @HallsofIvy, you're correct in [MATH] e^2[/MATH] being a typo - sorry about that.