1. ## Derivatives

Differentiate the following function:

$\displaystyle \frac{3x}{x^2+4}$

My attempt:
First I get rid of the fraction:
$\displaystyle f'(x) = (3x)(x^2+4)^-1$
Then I apply product rule F(x) = f(x)g(x) then F'(x) = f(x)g'(x)+f'(x)g(x):
$\displaystyle f'(x) = (3x)(x^2+4)'^-1 + (3x)'(x^2+4)^-1$
I differentiate the derivatives:
$\displaystyle (3x)(2x)^-1 + (3)(x^2+4)^-1$
I simplify:
$\displaystyle \frac{1}{6x^2} + \frac{1}{3x^2+12}$

This is the farthest I can get. What am I doing wrong? Thanks for all the help in advance.

2. Okay, that is already not the good way around. You haven't found the derivative yet, hence, it's f(x), not f'(x)

$\displaystyle f(x) = (3x)(x^2+4)^{-1}$

Then, since you already used f(x), you should use something else. Let's take u and v, such that:

$\displaystyle f(x) = uv$

$\displaystyle f'(x) = uv' + vu'$

$\displaystyle u = 3x$, $\displaystyle v = (x^2 + 4)^{-1}$

$\displaystyle u' = 3$, $\displaystyle v' = -(x^2+4)^{-2}.2x$

Now, substitute that into f'(x).

3. Originally Posted by Unknown008
Okay, that is already not the good way around. You haven't found the derivative yet, hence, it's f(x), not f'(x)

$\displaystyle f(x) = (3x)(x^2+4)^{-1}$

Then, since you already used f(x), you should use something else. Let's take u and v, such that:

$\displaystyle f(x) = uv$

$\displaystyle f'(x) = uv' + vu'$

$\displaystyle u = 3x$, $\displaystyle v = (x^2 + 4)^{-1}$

$\displaystyle u' = 3$, $\displaystyle v' = -(x^2+4)^{-2}.2x$

Now, substitute that into f'(x).
Is this a university level answer? Cause I'm only doing high-school calculus, and have no idea on how I would solve by using your method. Is it impossible to do it the method I was doing? Just when I thought I was getting the hang of it, too.

Thanks for all the help, by the way. ^^

4. No, I'm not in university, though I try to solve some problems from that level, or at least simplify.

In fact, I'm using your method. Maybe, I should have used f(x) and g(x). I'm just putting f(x), g(x), and their derivatives separately so that you see them more clearly.

Let $\displaystyle y = (3x)(x^2+4)^{-1}$

Then, I apply the rule:

If $\displaystyle y = f(x)g(x)$

Then

$\displaystyle y' = f(x)g'(x) + f'(x)g(x)$

From that, you know that $\displaystyle f(x) = 3x$ and $\displaystyle g(x) = (x^2 + 4)^{-1}$

You can find f'(x) and g'(x) form there?

Then, substitute the values in the rule:

$\displaystyle y' = f(x)g'(x) + f'(x)g(x)$

You already have f(x) and g(x):

$\displaystyle y' = 3xg'(x) + f'(x)(x^2 + 4)^{-1}$

5. There's always the Quotient Rule...

$\displaystyle \displaystyle \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\,\frac{du}{dx} - u\,\frac{dv}{dx}}{v^2}$.

6. Originally Posted by Unknown008
No, I'm not in university, though I try to solve some problems from that level, or at least simplify.

In fact, I'm using your method. Maybe, I should have used f(x) and g(x). I'm just putting f(x), g(x), and their derivatives separately so that you see them more clearly.

Let $\displaystyle y = (3x)(x^2+4)^{-1}$

Then, I apply the rule:

If $\displaystyle y = f(x)g(x)$

Then

$\displaystyle y' = f(x)g'(x) + f'(x)g(x)$

From that, you know that $\displaystyle f(x) = 3x$ and $\displaystyle g(x) = (x^2 + 4)^{-1}$

You can find f'(x) and g'(x) form there?

Then, substitute the values in the rule:

$\displaystyle y' = f(x)g'(x) + f'(x)g(x)$

You already have f(x) and g(x):

$\displaystyle y' = 3xg'(x) + f'(x)(x^2 + 4)^{-1}$
Okay, this is what I get:
After differentiating g'(x) I get: $\displaystyle 2x^{-1}$ and after differentiating f'(x) I get: $\displaystyle 3$

Then I substitute them in:
$\displaystyle F'(x) = (3x)(2x)^{-1} + (3)(x^2+4)^{-1} => \frac{3x}{2x} + \frac{3}{x^2+4}$. Do I just cross multiply thereafter and I'm done?

Gah, rational functions seems to have always been my downfall in mathematics.

7. No, g'(x) is wrong.

For a function of the form:

$\displaystyle y = (f(x))^n$

$\displaystyle y' = n(f(x))^{n-1}.f'(x)$

Which means,

$\displaystyle g'(x) = -1.(x^2 + 4)^{-1-1}.2x = -2x(x^2 + 4)^{-2}$

Then, you put it in F'(x) as you did. However, you don't cross multiply. You usually can simplify

8. Originally Posted by Prove It
There's always the Quotient Rule...

$\displaystyle \displaystyle \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\,\frac{du}{dx} - u\,\frac{dv}{dx}}{v^2}$.
Unfortunately, my calculus teacher is a really stubborn and stern. He prefers us to only use his method despite the fact that other methods are more widespread and easier for students. He'll give us 0 on tests and assignments for using "foreign methods" as he likes to call it.

9. I'm sure that this is the next method you'll learn about in your class, as soon as you're confortable with the product rule, which is what you just used

10. Originally Posted by Unknown008
No, g'(x) is wrong.

For a function of the form:

$\displaystyle y = (f(x))^n$

$\displaystyle y' = n(f(x))^{n-1}.f'(x)$

Which means,

$\displaystyle g'(x) = -1.(x^2 + 4)^{-1-1}.2x = -2x(x^2 + 4)^{-2}$

Then, you put it in F'(x) as you did. However, you don't cross multiply. You usually can simplify
Okay, thanks for all the help thus far unknown.

Just one question: How did you get the $\displaystyle -2x$ coefficient in front of $\displaystyle (x^2+4)$? Doesn't it stay as $\displaystyle -1$ since you subtracted -1 from -2 after you applied the law?

11. The it is -1 multiplied by 2x, giving -2x.

Now, the 2x is obtained when you differentiate g(x). You first 'lower the power down' next to the bracket, decrease the power by 1 and last but not least, you get the derivative of what is inside the brackets.

The derivative of x^2 + 4 is 2x.

12. Okay:

Now I substitute them in:
$\displaystyle F'(x) = (3x)(2x)^{-1} +(3)(-2(x^2+4))^{-2}$
Then I simplify:
$\displaystyle \frac{3x}{2x} +\frac{3}{(-2x^2-8)^2}$
Now if I were to simplify further, I get the wrong answer seeing as the answer in my book is much different.

13. From what you posted earlier, only g'(x) was wrong, hence, correcting to the correct g'(x), we get:

$\displaystyle F'(x) = (3x)(-2x(x^2+4)^{-2}) + (3)(x^2+4)^{-1}$

Then you try simplifying

14. Thanks.

This is what I got:
$\displaystyle \frac{3x^3+12x+12x^4+96x^2+192}{(-2x-8)^2(x^2+4)}$

I tried simplifying that and yielded the wrong answer. What have I done wrong?

15. Hm... not really, unfortunately.

$\displaystyle F'(x) = (3x)(-2x(x^2+4)^{-2}) + (3)(x^2+4)^{-1}$

$\displaystyle F'(x) = \dfrac{(3x)(-2x)}{(x^2+4)^2} + \dfrac{3}{x^2+4}$

There, what you can do is factor the 3/(x^2 + 4)^2:

$\displaystyle F'(x) = \dfrac{3}{(x^2 + 4)^2}\left(-2x^2 +(x^2+4)\right)$

$\displaystyle F'(x) = \dfrac{3}{(x^2 + 4)^2}\left(4-x^2\right)$

$\displaystyle F'(x) = \dfrac{3(4-x^2)}{(x^2 + 4)^2}$

$\displaystyle F'(x) = \dfrac{12-3x^2}{(x^2 + 4)^2}$

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