# Homework Help: Multiple calculus questions

1. Dec 4, 2012

### tahayassen

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

Can somebody show how you can get from the second step to the third step? I tried factoring x+1 from the numerator but it just made it even messier.

2)

Can somebody tell me what method/technique or what theorem I need to use in order to do this question?

3)

Can somebody tell me what method/technique or what theorem I need to use in order to do this question?

Last edited: Dec 4, 2012
2. Dec 4, 2012

### Ray Vickson

Show your work: what have you tried so far?

3. Dec 4, 2012

### SammyS

Staff Emeritus
How does the denominator in the third step compare to the denominator in the second step?

4. Dec 4, 2012

### tahayassen

One more question:

4)

$${ (\frac { 1 }{ 3 } x+\frac { 1 }{ 3 } { x }^{ -1 }+{ x }^{ -2 }) }^{ -1 }\\ =\frac { 1 }{ 3 } { x }^{ -1 }+\frac { 1 }{ 3 } { x }+{ x }^{ 2 }$$

That's correct, right? But if it were -2 instead of -1, you wouldn't be able to distribute the exponent right? You'd still be able to distribute the negative though, correct?

I'll upload my work in a sec

Last edited: Dec 4, 2012
5. Dec 4, 2012

### tahayassen

I solved question 1. The key was to use the product rule - not the quotient rule for derivatives and then use common denominators to add the fractions.

For question 2, I tried to use the discriminant for cubic functions to make some sort of resemblance to the inequality, but no luck. See: http://en.wikipedia.org/wiki/Discriminant#Formula_2

Here is my attempt for question 3:

$$\underset { x\rightarrow \infty }{ lim } \frac { { x }^{ 2011 }+{ 2010 }^{ x } }{ -{ x }^{ 2010 }+{ 2011 }^{ x } } \\ =\underset { x\rightarrow \infty }{ lim } \frac { { { x }^{ 2010 }(x }+{ 2010 }^{ x }{ x }^{ -2010 }) }{ { x }^{ 2010 }(-1+{ 2011 }^{ x }{ x }^{ -2010 }) } \\ =\underset { x\rightarrow \infty }{ lim } \frac { { x }+{ 2010 }^{ x }{ x }^{ -2010 } }{ -1+{ 2011 }^{ x }{ x }^{ -2010 } }$$

Just to add, I have a calculus exam in 4 hours and 15 minutes from the time of this post, so sorry about not showing my work the first time.

6. Dec 4, 2012

### Staff: Mentor

No, it's not. You can't distribute exponents over a sum or difference, and that seems to be what you did.

7. Dec 4, 2012

### Staff: Mentor

I don't think this will be any help. The dominant term is the 2011x in the denominator, which leads me to believe that the limit of the overall expression is 0. I would be inclined to use L'Hopital's rule to confirm that guess.

8. Dec 4, 2012

### tahayassen

I'm currently on my tablet, so I won't be able to post latex.

Doh. I'm not sure how that slipped from my mind. I'll try it in a second. I think this is the right way.

Thanks for answering question 4. So if instead of a negative exponent for the brackets, the contents of the brackets was just in the denominator of a fraction, then you still wouldn't be able to move each term to the numerator by changing the signs of the exponents right?

Anyone know how to solve question 2?

9. Dec 4, 2012

### SammyS

Staff Emeritus
You can easily get the desired answer after correcting your 'second line':

$\displaystyle f'(x)=\frac{2\sqrt { x+1 }}{2\sqrt { x+1 }}\left(\frac { 2x\sqrt { x+1 } -\frac {x^2 }{ 2\sqrt { x+1} } }{ x+1 }\right)$
$\displaystyle =\frac{4x(x+1)-x^2}{2(x+1)^{\frac{3}{2}}}$

$\displaystyle =\frac{x(3x+4)}{2(x+1)^{\frac{3}{2}}}$​

10. Dec 5, 2012

### tahayassen

The solution to #2 was to take the derivative of the equation. Then calculate the discriminant for the the derivative.

The solution to #3 was to apply L'Hôpital's rule 2011 times (or when you noticed the pattern and skipped right to the end).