Question

actually all I really need to know is how to find (x+h)^1.5

Answer 1

what do you actually actually need to do because there is nothing to "find" here. it is what it is

Answer 2

you can rewrite it as
\[(x+h)^{\frac{3}{2}}\] or as
\[\sqrt{(x+h)^3}\] or even as
\[(x+h)\sqrt{x+h}\] if you like

Answer 3

I am trying to find the derivative of f(x) = (6x)^1.5 - (4x)^.5

Answer 4

what you actually need is (x+h)^1.5= x+1.5*h*x^0.5+...O(h²)

Answer 5

then it is
\[9\sqrt{6x}-\frac{1}{\sqrt{x}}\]

Answer 6

and you see, I have the answer in the back of my book, which is (9x)^.5 - (2x) ^(-.5)

Answer 7

you want to use the limit method... dont you?

Answer 8

I just really don't understand how to get there

Answer 9

he didn't specify which method to use?

Answer 10

the way path of solving it using the lim h...0 (f(x+h)-f(x)/)h is horrible if you havent seen binomial expansions or infinite series, I can tell you the "trick" if you want

Answer 11

i will tell you what they used. they used the "power rule"

Answer 12

oh yes, okay the limit method..duh. because we are finding it as the limit is approaching 0. yes. that is what you meant by lim h....0

Answer 13

okay, what is the power rule?

Answer 14

the derivative of x^n= n*x^(n-1)

Answer 15

first off i am fairly sure that answer is wrong. lets check

Answer 16

\[(6x)^{\frac{3}{2}}=\sqrt{6^3}x^{\frac{3}{2}}=6\sqrt{6}x^{\frac{3}{2}}\]

Answer 17

so the derivative is
\[\frac{3}{2}\times 6\sqrt{6}x^{\frac{1}{2}}=9\sqrt{6}\sqrt{x}=9\sqrt{6x}\]

Answer 18

which is not the same as
\[\sqrt{9x}\]

Answer 19

second part is wrong too. what book is this?

Answer 20

Lial Hunderford Holcomb Mathematics with applications 9th ed

Answer 21

anyway , why would a book give an answer in this form:
\[\sqrt{9x}\]
and not
\[3\sqrt{x}\]

Answer 22

hmmm i have seen it. lial and hunferford used to be lial miller.
@fiddlearound good point, and it is wrong in any case

Answer 23

u sure it was (4x)^0.5 and not 4x^(0.5)

Answer 24

@scountz it is a dumb question with decimal exponents and the answer given is wrong, so forget about it. or impress your teachers and say "i don't think this is right"

Answer 25

oooh maybe it was
\[6x^{\frac{3}{2}}+4x^{\frac{1}{2}}\]in which case the answer is
\[9x^{\frac{1}{2}}+2x^{-\frac{1}{2}}\]

Answer 26

aka
\[9\sqrt{x}+\frac{2}{\sqrt{x}}\]

Answer 27

but it is certainly not
\[(9x)^{.5}\] for sure