The answer to question 1 is again rewritten as
I think.... But I've been known to be wrong
Some more questions, and then Im done with Calculus until Sunday or Monday!
The first question;
I rewrite it as
Then I apply The Power Rule and The Difference Rule once I have the derivatives and end up at
Am I missing anything?
The second question is
Which gets rewritten as
Which then I apply the product rule and get
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.
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.
Which leaves me with
And thats much different from what I tried getting the first go around.
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
$y = \dfrac{3x}{x^2 + 4} = \dfrac{u}{v} \implies \dfrac{dy}{dx} = \dfrac{\dfrac{du}{dx} * v - u * \dfrac{dv}{dx}}{v^2} = \dfrac{3v - u * 2x}{v^2} = \dfrac{3(x^2 + 4) - 3x * 2x}{v^2} = \dfrac{3x^2 + 12 - 6x^2}{v^2} = \dfrac{3(4 - x^2)}{(x^2 + 4)^2}.$
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 as v
After writing it out myself, it is clear, I had it just had to factor I believe. Thanks JEFF!!!!!!!!!
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.
Recap: . You wrote the derivative is .
Your error was not in factoring. It was in applying the Chain Rule.
Using the product rule, the derivative of is
After taking the derivative of , the first term becomes: . This matches your first term.
After taking the derivative of , the second term becomes . 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.