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

*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.

 

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]

 

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].

Almost. The numerator should be a difference (not a sum).

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.

 

Just a small error, it's 3x²h + 3xh² + h³ in the numerator (not h²).
And in the denominator it's x^{3 / 2}, not 3^{3 / 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,

 

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,