# anyone help me with a basic calculus problem?

Discussion in 'General Discussion' started by Sage, Sep 11, 2006.

1. ### Sage 13 short of a dozen Regular

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For some reason I cannot get this darned thing to work out right...

Differentiate:
Sq.Rt. (x + sq.rt. (x) )

It seems to me like it should be:

1 / [ 4 * sq.rt.(x + sq.rt. (x)) * sq.rt. (x) ]

BUT, calculator indicates:

[2 * sq.rt. (x) + 1] / [4 * sq.rt. (x) * sq.rt. {sq.rt. (x) + (sq.rt. (x) + 1) } ]

I cannot figure out what the heck i'm doing wrong. I'm using chain rule and differentiating whole function, to get 1 / [2 * sq.rt. {x + sq.rt. (x) } ] for the first function, then 1 / [2 * sq.rt. (x) ] for the second one, multiplying them to get the above result. I tried graphing them on the calculator and comparing the slope of the tangent line, as computed by my calculator, of the initial function at x=0.75 and it matches the value of what my calculator indicates if the correct derivative at x=0.75 but it does not match what my work indicates is correct.

Would greatly appreciate any help.

#1
2. ### K.I.L.E.R Retarded moron Veteran

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To calculate the 7th order PDE...

I got it:

g(x) = x + x^0.5
q(a) = a^0.5

g'(x) = 1 + 0.5*x^-0.5 = 1 + 1/(2*sqrt(x))
q'(a) = 0.5*a^-0.5 = 1/2*sqrt(a)

q'(a)*g'(x) = 1/2*sqrt(a) * 1 + 1/(2*sqrt(x))
= 1/2*sqrt(x + sqrt(x)) * (1 + 1/2*sqrt(x))

I love maths.
Want to play again?

#2
3. ### pcchen Moderator ModeratorVeteranSubscriber

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It should be 1 + 1/(2*sqrt(x)) for the second one.

#3
4. ### K.I.L.E.R Retarded moron Veteran

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I got it first.
I got it before you did.

#4
5. ### Sage 13 short of a dozen Regular

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thanks guys, I already figured it out with the help of my mommie

#5
6. ### K.I.L.E.R Retarded moron Veteran

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I wasted a page in my book for that.

#6
7. ### Sage 13 short of a dozen Regular

Joined:
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Messages:
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Location:
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Well, because I gave upon her helping me but then she called me back.

#7
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