# Thread: What to do next? (derivative)

1. ## What to do next? (derivative)

Hi, I'm beginner to calculus and I forgot my algebra because I stop my study for years. Now I don't know what next after I plug the value to the formula of quotient rule in derivate.

This is the given.
$y = \frac{x^3}{(x^2+4)^2}$

Then I think this is the derivative of the given function, but how to solve or factor or simplify this?
$= \frac{(x^2+4)^2(3x^2)-(x^3)2(x^2+4)(2x)}{((x^2+4)^2)^2}$

If you can show me the full solution then explain what process did you do, it will help me alot. Thanks

2. ## Re: What to do next? (derivative)

Originally Posted by avahdon
Hi, I'm beginner to calculus and I forgot my algebra because I stop my study for years. Now I don't know what next after I plug the value to the formula of quotient rule in derivate.

This is the given.
$y = \frac{x^3}{(x^2+4)^2}$

Then I think this is the derivative of the given function, but how to solve or factor or simplify this?
$= \frac{(x^2+4)^2(3x^2)-(x^3)2(x^2+4)(2x)}{((x^2+4)^2)^2}$

If you can show me the full solution then explain what process did you do, it will help me alot. Thanks
if

$f(x) = \dfrac{g(x)}{h(x)}$

and

$f^\prime(x)=\dfrac{df}{dx}(x)$

then

$f^\prime(x)=\dfrac{g^\prime(x)h(x)-g(x)h^\prime(x)}{(h(x))^2}$

so in this case

$g(x)=x^3$

$g^\prime(x)=3x^2$

$h(x)=(x^2+4)^2$

$h^\prime(x)=2(x^2+4)2x$

you should be able to put all the pieces together from here.

3. ## Re: What to do next? (derivative)

Originally Posted by romsek
if

$f(x) = \dfrac{g(x)}{h(x)}$

and

$f^\prime(x)=\dfrac{df}{dx}(x)$

then

$f^\prime(x)=\dfrac{g^\prime(x)h(x)-g(x)h^\prime(x)}{(h(x))^2}$

so in this case

$g(x)=x^3$

$g^\prime(x)=3x^2$

$h(x)=(x^2+4)^2$

$h^\prime(x)=2(x^2+4)2x$

you should be able to put all the pieces together from here.
I already solve the derivative and I have this $= \frac{(x^2+4)^2(3x^2)-(x^3)2(x^2+4)(2x)}{((x^2+4)^2)^2}$

my question now is how to simplify, solve, or factor this? because my teacher don't accept this answer

4. ## Re: What to do next? (derivative)

Originally Posted by avahdon
Hi, I'm beginner to calculus and I forgot my algebra because I stop my study for years. Now I don't know what next after I plug the value to the formula of quotient rule in derivate.

This is the given.
$y = \frac{x^3}{(x^2+4)^2}$

Then I think this is the derivative of the given function, but how to solve or factor or simplify this?
$= \frac{(x^2+4)^2(3x^2)-(x^3)2(x^2+4)(2x)}{((x^2+4)^2)^2}$

If you can show me the full solution then explain what process did you do, it will help me alot. Thanks
One thing you should be able to see that there is a term of the form $x^2+ 4$ in both terms in the numerator and one in the denominator so you can cancel those:
$\frac{(x^2+ 4)(3x^2)- (x^3)(2x)}{(x^2+ 4)^3}$
(You did recognize that $((x^2+ 4)^2)^2$ is $(x^2+ 4)^4$, right?)

Now multiply out $(x^2+ 4)(3x^3)$ and $(x^3)(2x)$ and subtract.

(I strongly recommend that you start reviewing algebra and trig.)

5. ## Re: What to do next? (derivative)

Hello, avahdon!

$y \:=\: \frac{x^3}{(x^2+4)^2}$

. . $y' \:=\: \frac{(x^2+4)^2(3x^2)-(x^3)2(x^2+4)(2x)}{((x^2+4)^2)^2}$

How to simplify this?

You have: . $y' \;=\;\frac{3x^2(x^2+4)^2 - 4x^4(x^2+4)}{(x^2+4)^4}$

Factor: . $y' \;=\;\frac{x^2(x^2+4)\big[3(x^2+4) - 4x^2\big]}{(x^2+4)^4}$

Reduce: . $y' \;=\;\frac{x^2\big[3x^2+12 - 4x^2\big]}{(x^2+4)^3}$

Simplify: . $y' \;=\;\frac{x^2(12-x^2)}{(x^2+4)^3}$

6. ## Re: What to do next? (derivative)

Originally Posted by HallsofIvy
One thing you should be able to see that there is a term of the form $x^2+ 4$ in both terms in the numerator and one in the denominator so you can cancel those:
$\frac{(x^2+ 4)(3x^2)- (x^3)(2x)}{(x^2+ 4)^3}$
(You did recognize that $((x^2+ 4)^2)^2$ is $(x^2+ 4)^4$, right?)

Now multiply out $(x^2+ 4)(3x^3)$ and $(x^3)(2x)$ and subtract.

(I strongly recommend that you start reviewing algebra and trig.)
thanks, can you tell me which topic should i review in algebra and trig?

Originally Posted by Soroban
Hello, avahdon!

You have: . $y' \;=\;\frac{3x^2(x^2+4)^2 - 4x^4(x^2+4)}{(x^2+4)^4}$

Factor: . $y' \;=\;\frac{x^2(x^2+4)\big[3(x^2+4) - 4x^2\big]}{(x^2+4)^4}$

Reduce: . $y' \;=\;\frac{x^2\big[3x^2+12 - 4x^2\big]}{(x^2+4)^3}$

Simplify: . $y' \;=\;\frac{x^2(12-x^2)}{(x^2+4)^3}$
thanks.

7. ## Re: What to do next? (derivative)

Alternatively, use logarithmic differentiation.
y=x^3/(x^2+4)^2, so
ln(y) = ln(x^3) - ln(x^2+4)^2 = 3ln(x)-2ln(x^2+4)
Differentiating both sides, you have
y'/y = 3/x - 2*1/(x^2+4)*2x = 3/x -4x/(x^2+4) = [3(x^2+4)-4x^2]/[x(x^2+4)] = [-x+12]/[x(x^2+4)]
Now, solving for y', you just multiply both sides by y:
y' = [-x+12]/[x(x^2+4)] * x^3/(x^2+4)^2 = (12-x)/(x^2+4) * x^2/(x^2+4)^2 = (12x^2-x^3)/(x^2+4)^3. //