- #1

NewScientist

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

Look forward to your replies

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- Thread starter NewScientist
- Start date

- #1

NewScientist

- 171

- 0

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

Look forward to your replies

- #2

quantumdude

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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

NewScientist

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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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Homework Helper

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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

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

quantumdude

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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

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

This thread really belongs in the Homework section, so I'm going to move it there.

Last edited:

- #6

NewScientist

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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

Homework Helper

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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

NewScientist

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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

Homework Helper

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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

NewScientist

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Look above I realisd I was being stupid!

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