Having trouble with a Calculus Question

[west_eh]

This is probably pretty easy but I'm stumped:

[ QUOTE ]
Use the formal definition of limit to find the derivative of f(x)=2/3x.

Begin by writing the formal definition (involving the limit) of f'(x).

[/ QUOTE ]

Here's what I've been doing:

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

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

=lim (2/3h(x+h))-(2/3xh)
h->0

=lim (2/h)(1/(x+h)-(1/x))
h->0

Then it's indeterminate and I dont know what to do.

Any help appreciated /images/graemlins/smile.gif.

[bunny]

[ QUOTE ]
This is probably pretty easy but I'm stumped:

[ QUOTE ]
Use the formal definition of limit to find the derivative of f(x)=2/3x.

Begin by writing the formal definition (involving the limit) of f'(x).

[/ QUOTE ]

Here's what I've been doing:

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

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

=lim (2/3h(x+h))-(2/3xh)
h->0

=lim (2/h)(1/(x+h)-(1/x))
h->0

Then it's indeterminate and I dont know what to do.

Any help appreciated /images/graemlins/smile.gif.

[/ QUOTE ]
f'(x) = lim [2/(3x+3h) - 2/(3x)].1/h

= lim 2/3.1/h[1/(x+h) - 1/x]

= lim 2/3.1/h[x/(x.x+h.x) - (x+h)/(x.x+h.x)]

= lim 2/3.1/h[-h/(x.x+h.x)]

= lim 2/3[-1/(x.x+h.x)]

= -2/3[1/x.x]

= -2/3x^2

[Hobbs.]

make a common demoninator.

[remission]

Are we looking for (2/3)x or 2/(3x)?

If it's (2/3)x then
f'(x)= lim h ->0 (2/3(x+h) -2/3x)/h
f'(x)= lim h ->0 (2/3x + 2/3h -2/3x)/h
f'(x)= lim h ->0 (2/3h)/h
f'(x)= lim h ->0 2/3

If it's 2/(3x)
f'(x)= lim h ->0 (2/(3x)-2/(3(x+h))/h
f'(x)= lim h ->0 (2/(3x)-2/(3x+3h))/h
-----Multiply 2/(3x) by (3x+3h)/(3x+3h)-----
f'(x)= lim h ->0 (2(3x+3h)/(3x)(3x + 3h)-2/(3x+3h))/h
-----Multiply 2/(3x+3h) by 3x/3x-----
f'(x)= lim h ->0 ((2(3x+3h)/(3x)(3x + 3h)-6x/(3x+3h)(3x))/h
f'(x)= lim h ->0 ((6x+6h)/(3x)(3x+3h)-6x/(3x+3h)(3x))/h
f'(x)= lim h ->0 (6h/(3x)(3x+3h))/h
f'(x)= lim h ->0 (6h/(3x^2+9xh))/h
f'(x)= lim h ->0 (6h/(3x^2+9xh) * (1/h)
f'(x)= lim h ->0 6/(3x^2+9xh)
----- Substitute 0 for h -----
f'(x)= lim h ->0 6/(3x^2+9x(0))
f'(x)= lim h ->0 6/(3x^2+0)
f'(x)= lim h ->0 3/2x^2

[Hobbs.]

you dropped a minus sign somewhere.

[west_eh]

Thanks alot for the replies