Solve Derivative of g(x) = x^(3/2) with First Principles

[forty]

[SOLVED] First principles

Determine the derivatives of the following function from first principles (i.e using the limit
definition of a derivative).

g(x) = x^(3/2)

lim h->0 : (x^(3/2) - (x + h)^(3/2) ) / (-h)

I understand first principles its more so the expansion of the (x + h)^(3/2) that has got me lost.

 

[Defennder]

Use the binomial theorem for non-integer powers. Then expand it for the first few terms, neglect powers of h by truncating the expansion (justified because h approaches zero) then substitute this result into the limit. Something will cancel out and you'll get the answer.

 

[forty]

(x + h)^(3/2)

=> ((x + h)^3))^(1/2)

(x^3 + 2hx^2 + x(h^2) + h(x^2) + 2x(h^2) + h^3)^(1/2)

So i get the following

(x^(3/2) - (x^3 + 2hx^2 + x(h^2) + h(x^2) + 2x(h^2) + h^3)^(1/2)) / (-h)

Firstly is this correct?
Secondly what do i do now? take out a factor form the square root, but what factor? >.<

 

[Gib Z]

The binomial theorem for non-integer powers is a bit of a deeper result than the derivative of a monomial isn't it?

Why not take this approach:

\frac{ (x+h)^{\frac{3}{2}} - x^{\frac{3}{2}} }{h} = \frac{ (x+h) \sqrt{x+h} - x\sqrt{x} }{h} = \frac{ x\sqrt{x+h} + h\sqrt{x+h} - x\sqrt{x}}{h}

Perhaps you can take it from here =]

 

[forty]

Thank you Gib Z!
As i said i was just having problems with the power 1.5 but after seeing it written like that made everything much clearer.

I multiplied by the conjugate and as a result canceled the h term from the denominator and got the proper result (3/2)x^(1/2).

Thanks!

 

[Gib Z]

Thank you Gib Z!
As i said i was just having problems with the power 1.5 but after seeing it written like that made everything much clearer.

I multiplied by the conjugate and as a result canceled the h term from the denominator and got the proper result (3/2)x^(1/2).

Thanks!

Good work, and no problem =] Just in case you are wondering, Defennder's suggestion works as well, though it may be beyond you at the moment.