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What is the derivative of x.(rootx) from first principles

  • #1
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The title is pretty self explanitary.

From first principles find f'(x) for f(x)=x.(rootx)

Look forward to your replies
 

Answers and Replies

  • #2
Tom Mattson
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*sniff sniff*

This smells a lot like homework to me!

If you want to calculate any derivative from first principles, you need to go back to the definition of the derivative (you can't get more first principled than that). If you recognize that [itex]x\sqrt{x}=x^{3/2}[/itex], you should be able to do it with a minimum of trouble.

Give it a shot and see if you can't do it.
 
  • #3
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f'(x) = f(a+h) + f(a)

f'(x) = lim (h->0) [(x+h)^3/2 + x^3/2]/h

how do you resolve the (x+h)^3/2)

?
 
  • #4
HallsofIvy
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NewScientist said:
f'(x) = f(a+h) + f(a)

f'(x) = lim (h->0) [(x+h)^3/2 + x^3/2]/h

how do you resolve the (x+h)^3/2)

?
Well first of all that's wrong: f'(x) is not f(a+h)- f(a).

In your second line, you at least have the limit but it's still wrong- remember the words difference quotient. You need a "difference" (so it is f(x+h)- f(x) not "+") and you need a "quotient"- divide by h.

[tex]f'(x)= lim_{h->0}\frac{(x+h)^{\frac{3}{2}}-x^{\frac{3}{2}}}{h}[/tex]

Try "rationalizing the numerator"- multiply both numerator and denominator by
[tex](x+h)^{\frac{3}{2}}+ x^{\frac{3}{2}}[/tex]
 
  • #5
Tom Mattson
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NewScientist said:
f'(x) = f(a+h) + f(a)
This could not possibly have come from a textbook. The definition of the derivative of a function is

[tex]f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}[/tex].

f'(x) = lim (h->0) [(x+h)^3/2 + x^3/2]/h
Almost. The numerator should be a difference (not a sum).

how do you resolve the (x+h)^3/2)

?
You rewrite the numerator correctly, and then you multiply the top and bottom of the difference quotient by the conjuagate of the numerator.

This thread really belongs in the Homework section, so I'm going to move it there.
 
Last edited:
  • #6
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I typed + instead of minus!!! The IT system was going down in 5 minutes so i rushed!

I get to (after rationalising the numerator) to

[tex]f'(x)=\lim_{h\rightarrow 0}\frac{(x+h)^3-x^3}{h[(x+h)^\frac{3}{2} + x^\frac{3}{2}]}[/tex].

[tex]f'(x)=\lim_{h\rightarrow 0}\frac{3x^2h+3h^2x +h^2}{h[(x+h)^\frac{3}{2} + x^\frac{3}{2}]}[/tex].

At this point i think I must have made an error for if the limit is taken you get 0/0 (all terms are multiplied by h).

Any idea?
 
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  • #7
VietDao29
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NewScientist said:
[tex]f'(x)=\lim_{h\rightarrow 0}\frac{3x^2h+3h^2x +h^2}{h[(x+h)^\frac{3}{2} + 3^\frac{3}{2}]}[/tex].
Any idea?
Just a small error, it's 3x2h + 3xh2 + h3 in the numerator (not h2).
And in the denominator it's x3 / 2, not 33 / 2
3x2h + 3xh2 + h3 = h(3x2 + 3xh + h2)
Note that the numerator and denominator have 'h' in common, so what can you do next?
Viet Dao,
 
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  • #8
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My error came because i had divided by h in my next line ofworking on paper but thought i may have gone wrong! dividing by h gives

[tex]f'(x)=\lim_{h\rightarrow 0}\frac{3x^2+3hx +h^2}{(x+h)^\frac{3}{2} + x^\frac{3}{2}}[/tex].

If i take the limit here I get

[tex]f'(x)=\frac{3x^2}{2(x^\frac{3}{2})}[/tex].

[tex]f'(x)=\frac{3}{2}x^\frac{1}{2}[/tex].

yey!
 
Last edited:
  • #9
VietDao29
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Yup, now take the limit as h tends to 0. It's no longer 0 / 0.
[tex]\lim_{h \rightarrow 0} \frac{3x ^ 2 + 3xh + h ^ 2}{(x + h) ^ \frac{3}{2} + x ^ \frac{3}{2}} = \frac{3x ^ 2 + 3x0 + 0 ^ 2}{(x + 0) ^ \frac{3}{2} + x ^ \frac{3}{2}}[/tex].
Can you go from here??
Viet Dao,
 
  • #10
167
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Look above I realisd I was being stupid!
 

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