Find the Derivative/Simplification of 2 Rational Expressions

[communitycoll]

Homework Statement

Find the derivative of:
y = [(2x - 5)^4][(8x^2 - 5)^-3]

Homework Equations

I get:
y' = -[(48x(2x - 5)^4) / (8x^2 - 5)^4] + [(8(2x - 5)^3) / (8x^2 - 5)^3]

Wolfram gets:
http://www.wolframalpha.com/input/?i=derivative+(2x+-+5)^4(8x^2+-+5)^-3

The Attempt at a Solution

I do everything Wolfram does here:
http://www.wolframalpha.com/input/?i=derivative+(2x+-+5)^4(8x^2+-+5)^-3
I just don't know how it simplifies what you see at the bottom (of the "Show Steps" section that is) to get the solution you see at the top.

 

[eumyang]

I do everything Wolfram does here:
http://www.wolframalpha.com/input/?i=derivative+(2x+-+5)^4(8x^2+-+5)^-3
I just don't know how it simplifies what you see at the bottom (of the "Show Steps" section that is) to get the solution you see at the top.

You should have learned how to add/subtract rational expressions in algebra. Just like in numerical fractions, you can only add/subtract if the denominators are the same. If they are not, you need to find the LCD (least common denominator) first. What is the LCD in this case?

 

[HallsofIvy]

Homework Statement

Find the derivative of:
y = [(2x - 5)^4][(8x^2 - 5)^-3]

Homework Equations

I get:
y' = -[(48x(2x - 5)^4) / (8x^2 - 5)^4] + [(8(2x - 5)^3) / (8x^2 - 5)^3]

So
$$y'= -\frac{48x(2x- 5)^4}{(8x^2- 5)^4}+ \frac{8(2x- 5)^3}{8x^2- 5)^3}$$

Wolfram gets:
http://www.wolframalpha.com/input/?i=derivative+(2x+-+5)^4(8x^2+-+5)^-3

The Attempt at a Solution

I do everything Wolfram does here:
http://www.wolframalpha.com/input/?i=derivative+(2x+-+5)^4(8x^2+-+5)^-3
I just don't know how it simplifies what you see at the bottom (of the "Show Steps" section that is) to get the solution you see at the top.

You have two fractions, one with denominator $(8x^2- 5)^4$, the other with denominator $(8x^2- 5)^3$. You can first simplify the calculation by factoring $(2x- 5)^3$ out of the numerators and $(8x^2- 5)^3$ out of the denominators:
$$\frac{(2x-5)^3}{(8x^2-5)^3}\left(\frac{-48x(2x-5)}{8x^2-5}+ 4\right)$$
Now, get a common denominator by multiplying that "4" by $(8x^2-5)/(8x^2- 5)$.
$$\frac{(2x-5)^3}{(8x^2- 5)^3}\left(\frac{-48(x(2x-5)}{8x^2- 5}+ \frac{4(8x^2- 5}{8x^2- 5}\right)$$
$$=\frac{(2x-5)^3}{(8x^2- 5)^3}\left(\frac{-96x^2+ 240+ 32x^2- 20}{8x^2- 5}\right)$$
$$= \frac{(2x-5)^3}{(8x^2- 5)^3}\left(\frac{220- 64x^2}{8x^2- 5}\right)$$

 

[communitycoll]

Okay then. I understand. Thank you both.