1. More Calculus Derivatives

Some more questions, and then Im done with Calculus until Sunday or Monday!

The first question;

$\displaystyle \sqrt{x}-\sqrt[3]{x}$

I rewrite it as $\displaystyle x^{\frac{1}{2}}-x^{\frac{1}{3}}$

Then I apply The Power Rule and The Difference Rule once I have the derivatives and end up at

$\displaystyle \frac{1}{2}x^{\frac{-1}{2}}-\frac{1}{3}x^{\frac{-2}{3}}$

Am I missing anything?

The second question is $\displaystyle \frac{3x}{x^{2}+4}$

Which gets rewritten as $\displaystyle (3x)(x^{2}+4)^{-1}$

Which then I apply the product rule and get

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

Correct?

2. Re: More Calculus Derivatives

The answer to question 1 is again rewritten as $\displaystyle \frac{1}{\sqrt{\frac{1}{2}}{x}}-\frac{2}{\sqrt[3]{\frac{1}{3}}{x}}$

I think.... But I've been known to be wrong

3. Re: More Calculus Derivatives

Originally Posted by rhickman
Some more questions, and then Im done with Calculus until Sunday or Monday!

The first question;

$\displaystyle \sqrt{x}-\sqrt[3]{x}$

I rewrite it as $\displaystyle x^{\frac{1}{2}}-x^{\frac{1}{3}}$

Then I apply The Power Rule and The Difference Rule once I have the derivatives and end up at

$\displaystyle \frac{1}{2}x^{\frac{-1}{2}}-\frac{1}{3}x^{\frac{-2}{3}}$

Am I missing anything?

The second question is $\displaystyle \frac{3x}{x^{2}+4}$

Which gets rewritten as $\displaystyle (3x)(x^{2}+4)^{-1}$

Which then I apply the product rule and get

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

Correct?
The first answer is correct. As a matter of presentation, however, I'd convert back to radicals.

$\dfrac{1}{2} * x^{-(1/2)} - \dfrac{1}{3} * x^{(-2/3)} = \dfrac{1}{2x^{(1/2)}} -\dfrac{1}{3x^{(2/3)}} = \dfrac{1}{2\sqrt{x}} - \dfrac{1}{3\sqrt[3]{x^2}}.$

As for the second question, I can't figure out where you went of the rails because you do not show your work. You can do this derivative using the product and power rules or using the quotient rule. You need to take another go at this one.

EDIT: While I was writing my previous post, you posted your conversion to radicals. You may need to review that notation.

4. Re: More Calculus Derivatives

Originally Posted by JeffM
The first answer is correct. As a matter of presentation, however, I'd convert back to radicals.

$\dfrac{1}{2} * x^{-(1/2)} - \dfrac{1}{3} * x^{(-2/3)} = \dfrac{1}{2x^{(1/2)}} -\dfrac{1}{3x^{(2/3)}} = \dfrac{1}{2\sqrt{x}} - \dfrac{1}{3\sqrt[3]{x^2}}.$

As for the second question, I can't figure out where you went of the rails because you do not show your work. You can do this derivative using the product and power rules or using the quotient rule. You need to take another go at this one.

EDIT: While I was writing my previous post, you posted your conversion to radicals. You may need to review that notation.
I will definitely be looking over my conversion to radicals, I see where I went wrong though, and I thought I had shown my work for the second question.... I tried using the product rule, I dont think they want me to simplify from there, which is where I left it.

5. Re: More Calculus Derivatives

For question 2, There aren't really any steps I can show, I haven't learned the quotient rule but I did find it on the internet, and tried to use it but couldn't quite grasp it.

When I tried the quotient rule I ended up with.

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

Which leaves me with $\displaystyle \frac{-3x^{2}+12}{(x^{2}+4)^{2}}$

And thats much different from what I tried getting the first go around.

6. Re: More Calculus Derivatives

I am going to show you how to do your second derivative TWO ways: using product, power, and chain rules and using the quotient rule (which may be easier for this problem).

$y = \dfrac{3x}{x^2 + 4}.$

$Let\ u = 3x \implies \dfrac{du}{dx} = 3.$

$Let\ v = x^2 + 4 \implies \dfrac{dv}{dx} = 2x.$

Product, Power, and Chain Rules

$Let\ w = v^{-1} \implies \dfrac{dw}{dv} = - 1 * v^{-2} = \dfrac{- 1}{v^2} \implies \dfrac{dw}{dx} = \dfrac{dw}{dv} * \dfrac{dv}{dx} = \dfrac{-1}{v^2} * 2x = \dfrac{-2x}{v^2} = \dfrac{-2x}{(x^2 + 4)^2}.$

$Also\ w = v^{-1} = \dfrac{1}{v} = \dfrac{1}{x^2 + 4}.$

$y = \dfrac{3x}{x^2 + 4} = \dfrac{u}{v} = u * v^{-1} = uw.$

$\therefore \dfrac{dy}{dx} = \dfrac{du}{dx} * w + u * \dfrac{dw}{dx} = 3 * w + u * \dfrac{-2x}{(x^2 + 4)^2} = 3 * \dfrac{1}{x^2 + 4} + 3x * \dfrac{-2x}{(x^2 + 4)^2} = \dfrac{3}{(x^2 + 4)} + \dfrac{- 6x^2}{(x^2 + 4)^2} =$

$\dfrac{3(x^2 + 4)}{(x^2 + 4)^2} + \dfrac{- 6x^2}{(x^2 + 4)^2} = \dfrac{3x^2 + 12 - 6x^2}{(x^2 + 4)^2} = \dfrac{3(4 - x^2)}{(x^2 + 4)^2}.$

Quotient Rule

$y = \dfrac{3x}{x^2 + 4} = \dfrac{u}{v} \implies \dfrac{dy}{dx} = \dfrac{\dfrac{du}{dx} * v - u * \dfrac{dv}{dx}}{v^2} = \dfrac{3v - u * 2x}{v^2} = \dfrac{3(x^2 + 4) - 3x * 2x}{v^2} = \dfrac{3x^2 + 12 - 6x^2}{v^2} = \dfrac{3(4 - x^2)}{(x^2 + 4)^2}.$

7. Re: More Calculus Derivatives

So I almost had it, had I factored out -3, but that would've still left it incorrect.

Im going to go over this Jeff. Thanks a lot.

Does that equation in the quotient portion mean you assigned 3x as u and $\displaystyle x^{2}+4$ as v

After writing it out myself, it is clear, I had it just had to factor I believe. Thanks JEFF!!!!!!!!!

8. Re: More Calculus Derivatives

That's why I set up u and v and their derivatives before showing both ways. $u = 3x\ and\ v = x^2 + 4$ in both derivations to keep things consistent.

Actually, if I could give a bit of advice. When you are just starting off, don't skip steps. Yes, you will save time, but you will make many errors.

9. Re: More Calculus Derivatives

Ok, thats a good point, thanks again for the help it's really appreciated.

10. Re: More Calculus Derivatives

Originally Posted by rhickman
Ok, thats a good point, thanks again for the help it's really appreciated.
Recap: $\displaystyle f(x) = \dfrac{3x}{x^2+4} = (3x)(x^2+4)^{-1}$. You wrote the derivative is $\displaystyle (3)(x^2+4)^{-1}+(3x)(-1)(2x)^{-2}$.
Your error was not in factoring. It was in applying the Chain Rule.

Using the product rule, the derivative of $\displaystyle (3x)(x^2+4)^{-1}$ is $\displaystyle \left[\dfrac{d(3x)}{dx}\right](x^2+4)^{-1} + (3x)\left[\dfrac{d(x^2+4)^{-1}}{dx}\right]$

After taking the derivative of $\displaystyle 3x$, the first term becomes: $\displaystyle (3)(x^2+4)^{-1}$. This matches your first term.

After taking the derivative of $\displaystyle (x^2+4)^{-1}$, the second term becomes $\displaystyle (3x)(-1)(x^2+4)^{-2}(2x)$. That does not match your second term at all. Your error had nothing to do with factoring. You did not apply the Chain Rule correctly.