Question

Derivative of f(x) = sqrt(x) - 5/sqrt(x) + 5x^7/3+3-2/X^3.

Answer 1

the square roots giving you a little trouble? :o

Answer 2

\[f(x)=\sqrt{x} - \frac{ 5 }{ \sqrt{x} } + 5x ^{\frac{ 7 }{ 3 }} + 3 + \frac{ 2 }{ x^{3} }\]

Math Syntax.

Answer 3

So far I understand:
\[\sqrt{x} = \frac{ 1 }{ 2 }x ^{\frac{ -1 }{ 2 }}\]

Answer 4

-2/x^3 < Correction.

Answer 5

hmmm looks good, so how about the second term?
\[-5x^\frac{ -1 }{ 2 }\]

The important thing to keep in mind with the power rule is:
When you're decreasing negative powers, they're actually increasing in the negative direction.

Answer 6

\[(d/dx)x^{1/2}=(1/2)x^{-1/2}\]
like you said the first time.

Answer 7

Oh I mean for the last bit after the stand alone 3 should be minues not plus. The math syntax post had that typo.

Answer 8

oh oh oh XD

Answer 9

lemme give you the last term, and maybe it'll help ya figure out the second term :O
\[(d/dx)\frac{ -2 }{ x^3 }=(d/dx)-2x^{-3}=(-2)(-3)x^{-4}\]
make sense? :o

Answer 10

Taking the reciprocals?

Answer 11

\[\frac{ 1 }{ x^3 }=x^{-3}\]
is that the part that's confusing you? :o
Yah you bring the x into the numerator, and apply a negative exponent to it.

Answer 12

Is it then -5x^2/3 ?

Answer 13

-2/3 That is.

Answer 14

-3/2 I mean gah.

Answer 15

Considering a -1/2 - 1 = -3/2.

Answer 16

\[(d/dx)\frac{ -5 }{ \sqrt x }=(d/dx)-5x^{-1/2}=(-5)(-1/2)x^{-3/2}\]
Yah your exponent looks correct :)

Answer 17

I'm going to follow up with my answer soon.

Answer 18

Ok, so I went for a meal break, Would this be correct:\[\frac{ 1 }{ 2 }x ^{\frac{ -1 }{ 2 }} - (-5)(\frac{ 1 }{ 2 })x ^{\frac{ -3 }{ 2 }} + 5(\frac{ 7 }{ 3 })x ^{\frac{ 4 }{ 3 }}-(-2)(-3)x ^{-4}\]

Which then after changing signs:
\[\frac{ 1 }{ 2 }x ^{\frac{ -1 }{ 2 }} + 5(\frac{ 1 }{ 2 })x ^{\frac{ -3 }{ 2 }} + 5(\frac{ 7 }{ 3 })x ^{\frac{ 4 }{ 3 }} +2(-3)x ^{-4}\]

?

Answer 19

And I assume taking the derivative of the +3 gave you 0?
Yah looks correct :) good job.

Answer 20

Yep, constants = 0 so +/- 0 just eliminates the 3. Thank you so much. I feel this forum is really great for help. I am going to contribute to others hopefully as much as I learn from others. :D