How Do You Simplify Complex Derivatives in Calculus?

[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?

 

[Ray Vickson]

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 { 3(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?

Show your work: what have you tried so far?

 

[SammyS]

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 { 3(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.

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

 

[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

 

[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.

 

[Mark44]

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?

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

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

 

[Mark44]

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 } }

I don't think this will be any help. The dominant term is the 2011^x 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.

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.

 

[tahayassen]

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

I don't think this will be any help. The dominant term is the 2011^x 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.

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?

 

[SammyS]

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.

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 <br /> =\frac{4x(x+1)-x^2}{2(x+1)^{\frac{3}{2}}}

\displaystyle <br /> =\frac{x(3x+4)}{2(x+1)^{\frac{3}{2}}}​

 

[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).