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Multiple calculus questions 
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#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

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#3
Dec412, 11:17 AM

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#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

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#7
Dec412, 01:59 PM

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#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

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[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). 


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