Simple Single Variable Derivative Question

[cappygal]

wow! I am having a major brain fart here .. I'm in multivariable calc but I need to take the derivative of a single variable function using the definition of derivative. The function is:

f(x)=abs(x)^(3/2) and I need to find the derivative when a=0

so by the def of derivative, the derivative equals:

lim (h->0) of [abs(a+h)^(3/2)-abs(a)^3/2]/h

I know I'm not allowed to just "plug in" a=0 to get the derivative at this point .. but how do I do it? I can't combine the absolute value terms because of the triangle inequality (for ex. abs(-6)-abs(3)=3 but abs(-6-3)=9)

help! I feel really stupid right now.

 

[quasar987]

why can't you plug a=0 ?

 

[cappygal]

I thought when you were using definition of derivitive, you aren't allowed to plug in for a until you have it simplified? Am I wrong? I could be .. it's been a year since I've had to work with definition of derivative.

 

[hypermorphism]

There's no such rule. "a" is a placeholder for a constant; place your constant in there to get the derivative there. Leave it as "a" to get an expression for the derivative (if it exists) at all values of "a".

 

[benorin]

wow! I am having a major brain fart here .. I'm in multivariable calc but I need to take the derivative of a single variable function using the definition of derivative. The function is:
f(x)=abs(x)^(3/2) and I need to find the derivative when a=0
so by the def of derivative, the derivative equals:
lim (h->0) of [abs(a+h)^(3/2)-abs(a)^3/2]/h
I know I'm not allowed to just "plug in" a=0 to get the derivative at this point .. but how do I do it? I can't combine the absolute value terms because of the triangle inequality (for ex. abs(-6)-abs(3)=3 but abs(-6-3)=9)
help! I feel really stupid right now.

use the *other* definition of limit:
f '(a) = lim (x->a) of [abs(x)^(3/2)-abs(a)^3/2]/(x-a)

then factor the top as a difference of cubes.

 

[HallsofIvy]

wow! I am having a major brain fart here .. I'm in multivariable calc but I need to take the derivative of a single variable function using the definition of derivative. The function is:
f(x)=abs(x)^(3/2) and I need to find the derivative when a=0
so by the def of derivative, the derivative equals:
lim (h->0) of [abs(a+h)^(3/2)-abs(a)^3/2]/h
I know I'm not allowed to just "plug in" a=0 to get the derivative at this point .. but how do I do it? I can't combine the absolute value terms because of the triangle inequality (for ex. abs(-6)-abs(3)=3 but abs(-6-3)=9)
help! I feel really stupid right now.

You cannot plug in h= 0 because that would give 0/0. But if you are finding the derivative at a given x= a, you certainly can plug in that value of a before you start. You are looking for
lim_{h\rightarrow0}\frac{h^{\frac{3}{2}}}{h}
That should be easy.