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Full Version: How to evaluate f'' at a specific point?
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Dumb question...how in home or CAS do I enter f''(3 root (x-2)) where x=some number of interest?
Did you try:

(x-2)^(1/3)''

or:

∂((x-2)^(1/3),x)'
In trying to get the second derivative at a specified point, I'm having difficulty in accessing the proper syntax. The template gives me d/dx. I can then get, say, ((d 3 root x-2 / dx)/dx))|x=1.99.
This returns 0 instead of 478+. What am I doing wrong?
I would just do

diff(diff(3ROOT(x-2),x),x) and then
Ans|x=1.99 (Ans actually has to be copied explicitly from the command line above)

returns 478.76 . . . for me.

Or do you have to do it with the d/dx template?

[You can also define f(x):=3Root(x-2), and then simply type f''(1.99) also returns 478.76 . . .]
That works (thanks!). Seems like there should be a more direct way to do this.
(∂(∂((x-2)^(1/3),x),x)|(x )= 1.99)
Credit: Eddie's Math and Calculator Blog series, by Edward Shore
3.24.
More natural syntax:
f(x):=(2-x)^(1/3);
f''(1.99)
For a large number of derivatives, for example 5, run diff(f(x),x,5)(x=1.99)
Thanks...this is what I hoped for! And when using the n root entry, negative results are produced for this equation directly as opposed to a complex number result!
Can this syntax be used in the function graphing app...the diff(f(x),x,2) form? What is the proper syntax for this?
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