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

Look forward to your replies

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- #1

- 167

- 0

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

Look forward to your replies

- #2

Tom Mattson

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This smells a lot like homework to me!

If you want to calculate

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)

?

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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Well first of all that's wrong: f'(x) isNewScientist 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)

?

In your second line, you at least have the limit but it's still wrong- remember the words

[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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This could not possibly have come from a textbook. The definition of the derivative of a function isNewScientist said:f'(x) = f(a+h) + f(a)

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

Almost. The numerator should be af'(x) = lim (h->0) [(x+h)^3/2 + x^3/2]/h

You rewrite the numerator correctly, and then you multiply the top and bottom of the difference quotient by thehow do you resolve the (x+h)^3/2)

?

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?

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?

Last edited:

- #7

VietDao29

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Just a small error, it's 3xNewScientist 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?

And in the denominator it's

3x

Note that the numerator and denominator have 'h' in common, so what can you do next?

Viet Dao,

Last edited:

- #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!

[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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[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

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Look above I realisd I was being stupid!

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