
#1
Dec412, 08:54 AM

P: 273

1) [tex]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 } } } [/tex]
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? 



#2
Dec412, 10:37 AM

HW Helper
Thanks
P: 4,670





#3
Dec412, 11:17 AM

Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,395





#4
Dec412, 01:04 PM

P: 273

Multiple calculus questions
One more question:
4) [tex]{ (\frac { 1 }{ 3 } x+\frac { 1 }{ 3 } { x }^{ 1 }+{ x }^{ 2 }) }^{ 1 }\\ =\frac { 1 }{ 3 } { x }^{ 1 }+\frac { 1 }{ 3 } { x }+{ x }^{ 2 }[/tex] 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 



#5
Dec412, 01:41 PM

P: 273

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: [tex]\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 } } [/tex] 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
Dec412, 01:47 PM

Mentor
P: 20,961





#7
Dec412, 01:59 PM

Mentor
P: 20,961





#8
Dec412, 02:10 PM

P: 273

I'm currently on my tablet, so I won't be able to post latex.
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
Dec412, 02:13 PM

Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,395

[itex]\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) [/itex] [itex]\displaystyle 



#10
Dec512, 01:59 AM

P: 273

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). 


Register to reply 
Related Discussions  
Calculus Readiness Multiple Choice Test Answer Key Wrong?  Precalculus Mathematics Homework  12  
Multiple choice questions  Introductory Physics Homework  1  
multiple choice questions about SHM  Introductory Physics Homework  4  
Multiple Circuit Questions  Introductory Physics Homework  1  
Multiple Choice Questions ( NEED HELP!! )  Introductory Physics Homework  3 