Tough Problem

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ninguen

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Jan 1, 2012
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I've been working on this one for a while, and I can't figure out the logic to get the desired answer.

According to Shaums, I can take the derrivative of:

f(x) = [(x- 1)(x + 1)^(-1)]^(1/2)

And the answer should come to:

f'(x) = 1/[(x + 1)(x^2 - 1)^(1/2)]

Can anyone show me the steps they used to get this answer?
 
Let's work the other way. You show the steps and someone will help you get it sorted.
 
JeffM said:
Based on your prior post, you are studying algebra and were confused by mixed numbers. Now you post a problem in differential calculus.

What are you studying?
Are you in school or self-teaching?

It is very hard for us to give answers that help YOU if we have no information about you.

If you are studying beginning algebra, I suspect it will be very hard to explain a derivative to you. How did you jump from algebra to calculus in one day?
Click to expand...

I'm an older guy self-teaching calculus because I want to understand the universe better. I'm not in school, and I don't have access to other students or teachers.

I've been through the Khan Academy, and now I'm practicing my skills with problems from Shaum's. Many of these problems are much harrier than anything I've seen on YouTube. I'll get the answer part-way, and then they just jump to some bizarre conclusion where I have absolutely no idea how they got there.

As for my previous post... I know algebra. It's just that that particular notation caught me off-guard. I've gone back and revised my algebra, precalc and trig because I thought it would help me with the Shaum's stuff... but it hasn't helped me much.
 
Hello, ninguen!

According to Shaums, I can take the derrivative of: .\(\displaystyle f(x) \:=\: \big[(x- 1)(x + 1)^{-1}\big]^{\frac{1}{2}}\)

and the answer should come to: .\(\displaystyle f'(x) \:=\: \dfrac{1}{(x + 1)(x^2 - 1)^{\frac{1}{2}}}\)

Can anyone show me the steps they used to get this answer?
Click to expand...

I would rewrite the function: .\(\displaystyle f(x) \:=\:\left(\dfrac{x-1}{x+1}\right)^{\frac{1}{2}} \;=\;\dfrac{(x-1)^{\frac{1}{2}}}{(x+1)^{\frac{1}{2}}} \)

Differentiate: .\(\displaystyle f'(x) \;=\;\dfrac{(x+1)^{\frac{1}{2}}\cdot\frac{1}{2}(x-1)^{-\frac{1}{2}} - (x-1)^{\frac{1}{2}}\cdot\frac{1}{2}(x+1)^{-\frac{1}{2}}}{x+1} \)

. . . . Factor: .\(\displaystyle f'(x) \;=\;\dfrac{\frac{1}{2}(x-1)^{-\frac{1}{2}}(x+1)^{-\frac{1}{2}}\big[(x+1) - (x-1)\big]}{x+1} \)

. . . Simplify: .\(\displaystyle f'(x) \;=\;\dfrac{1}{2}\cdot\dfrac{1}{(x-1)^{\frac{1}{2}}}\cdot\dfrac{1}{(x+1)^{\frac{1}{2}}}\cdot\dfrac{2}{x+1}\)

. . . . . . . . . . \(\displaystyle f'(x) \;=\;\dfrac{1}{\big[(x-1)(x+1)\big]^{\frac{1}{2}}(x+1)} \)

. . . . . . . . . . \(\displaystyle f'(x) \;=\;\dfrac{1}{(x^2-1)^{\frac{1}{2}}(x+1)} \)
 
ninguen said:
I'm an older guy self-teaching calculus because I want to understand the universe better. I'm not in school, and I don't have access to other students or teachers.

I've been through the Khan Academy, and now I'm practicing my skills with problems from Shaum's. Many of these problems are much harrier than anything I've seen on YouTube. I'll get the answer part-way, and then they just jump to some bizarre conclusion where I have absolutely no idea how they got there.

As for my previous post... I know algebra. It's just that that particular notation caught me off-guard. I've gone back and revised my algebra, precalc and trig because I thought it would help me with the Shaum's stuff... but it hasn't helped me much.
Click to expand...

If really want to learn calculus - self-teaching could be very hard!!

Go to your local community college and register for in-class courses (do not take on-line course for fundamental subjects like Calculus).
 
ninguen said:
I've been working on this one for a while, and I can't figure out the logic to get the desired answer.

According to Shaums, I can take the derrivative of:

f(x) = [(x- 1)(x + 1)^(-1)]^(1/2)

And the answer should come to:

f'(x) = 1/[(x + 1)(x^2 - 1)^(1/2)]

Can anyone show me the steps they used to get this answer?
Click to expand...

Do the substitution x=x+1, sop derivatives equal ( 1st, 2nd....)

f(x) = x**(0.5) * (x+2)**(-0.5)

= (x/(x+2))**(0.5) = Z(x)

x**(0.5)=Z(x)*(x+2)**(0.5)

Now derive

(0.5)x**(-0.5)=Z*(x+2)**(0.5) + Zprime * (x+2)**(0.5)

(0.5)x**-0.5 - (x+2)**(0.5)* ( Z= x**0.5/(x+2)**-(0.5) ) * (x+2)**(0.5)

divide both sides by (0.5)x**(-0.5)

fprime = ___________________


fprime = 0.5* (x**2 -1 )**-0.5 * (x-1)/x

=(0.5)(x**2 - 1)**-0.5 *(x-1)/x
 
burakaltr said:
Do the substitution x=x+1, sop derivatives equal ( 1st, 2nd....)

f(x) = x**(0.5) * (x+2)**(-0.5)

= (x/(x+2))**(0.5) = Z(x)

x**(0.5)=Z(x)*(x+2)**(0.5)

Now derive

(0.5)x**(-0.5)=Z*(x+2)**(0.5) + Zprime * (x+2)**(0.5)

(0.5)x**-0.5 - (x+2)**(0.5)* ( Z= x**0.5/(x+2)**-(0.5) ) * (x+2)**(0.5)

divide both sides by (0.5)x**(-0.5)

fprime = ___________________


fprime = 0.5* (x**2 -1 )**-0.5 * (x-1)/x

=(0.5)(x**2 - 1)**-0.5 *(x-1)/x
Click to expand...

Yuck. I appreciate your wanting to help someone, but please learn Latex when you show so much detail like above.
 
@soroban

Maybe I've been doing it wrong... this is what I get for the first part

f(x) = (x-1)^(1/2) x (x+1)^(-1/2)

I turn both top and bottom into square roots, then do the product rule.

D[(x-1)^(1/2)] x (x+1)^(-1/2) + (x-1)^(1/2) x D[(x+1)^(-1/2)]


And that leaves me with the following result:


f'(x) = (1/2) x (1) x (x-1)^(-1/2) x (x+1)^(-1/2) + (x-1)^(1/2) x (1/2) x (1) x (x+1)^(-3/2)

I think I might be messing up because I get that -3/2 exponent when I decrement the exponent by one at the end. Is there a mistake that I'm making here? It looks nothing like what you got.

@Subhotosh

Right now, my life and schedule doesn't allow for taking a class. (Otherwise, I wouldn't be spending all my time typing out questions on internet message boards) I'm disciplined, patient and focused... and I'm not in a rush to beat an exam. So I'm hoping that you're wrong about the self-teaching.
 
Last edited:
Maybe I've been doing it wrong... this is what I get for the first part

f(x) = (x-1)^(1/2) x (x+1)^(-1/2)

I turn both top and bottom into square roots, then do the product rule.

D[(x-1)^(1/2)] x (x+1)^(-1/2) + (x-1)^(1/2) x D[(x+1)^(-1/2)]


And that leaves me with the following result:


f'(x) = (1/2) x (1) x (x-1)^(-1/2) x (x+1)^(-1/2) + (x-1)^(1/2) x (1/2) x (1) x (x+1)^(-3/2)

I think I might be messing up because I get that -3/2 exponent when I decrement the exponent by one at the end. Is there a mistake that I'm making here?
 
Oops. Careless mistaks.

Thank you all so much for your help. I understand this now.
 
ninguen said:
For extremely small values of 1
Click to expand...

Water = Water + a very little tiny bit of cyanide.

Drink the potion on the right hand side of the equal sign. After all, it's EXACTLY like the pure water on the left hand side. ;)
 
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