Question

Find the derivative of g(x)= -3/ squareroot x

Answer 1

\[g(x)=\frac{-3}{\sqrt{x}}=\frac{-3}{x^{\frac{1}{2}}}=-3x^{\frac{-1}{2}}\]Now you can just use the power rule.

Answer 2

\[\frac{d}{dx}x^{n}=n\times x^{n-1}\]

Answer 3

for petes sake to not use the product rule with constants

Answer 4

quotient rule then ?

Answer 5

no
power rule

Answer 6

@Aylin wrote the method above

Answer 7

Power rule, Aylin have posted it up here.

Answer 8

\[y'= -\frac{ 3 }{ 2 }x^{-\frac{ 1 }{ 2 }}\]

Answer 9

like that @satellite73 @Aylin

Answer 10

You didn't execute the power rule properly.

Answer 11

\[g(x) = -3x ^{\frac{ -1 }{ 2 }}\]g'(x) will be:
\[g'(x)=(-3*-\frac{ 1 }{ 2 })*x ^{-\frac{ 1 }{ 2 }-1}=\frac{ 3 }{ 2 } *x^{\frac{ -3 }{ 2 }} \iff g'(x)=\frac{ 3 }{ 2\sqrt{x^3} }\]

Answer 12

Okay but the thing is you use the power rule first and then take the derivative ?:S

Answer 13

No, the power rule is a way to differentiate. :]

Answer 14

@zairhenrique then why is it done twice here ?

Answer 15

It's only done once

Answer 16

I am really confused

Answer 17

the "power rule" is not the same as "write in exponential form"
they are two different things

Answer 18

for example, if you want the derivative of \(f(x)=\frac{2}{x^3}\) then you can REWRITE it as
\[f(x)=2x^{-3}\] and then use the power rule to get
\[f'(x)=-3\times 2x^{-3-1}=-6x^{-4}\]

Answer 19

\[g(x) = \frac{ -3 }{ \sqrt{x}} \]

Answer 20

that is not "using the power rule twice" that is
"write in exponential form" then "use the power rule"

Answer 21

\[g(x)=-3x^{\frac{ 1 }{ 2 }}\]

Answer 22

writing
\[g(x)=-3x^{-\frac{1}{2}}\] is not the power rule, it is getting it in the form so that you can use the power rule

Answer 23

Yes i know

Answer 24

btw
\[\frac{3}{\sqrt{x}}\neq -3x^{\frac{1}{2}}\]it is
\[-3x^{-\frac{1}{2}}\]

Answer 25

ohh i see the problem
make sure to write it in exponential form correctly

Answer 26

\[g'(x)=\frac{ -3 }{ 2 }x^\frac{ 1 }{ 2 }\]

Answer 27

ops that exponent is supposed ot be negative

Answer 28

\[= \frac{ -3 }{ 2\sqrt{x} }\]

Answer 29

It's not right... The -3/2 became positive and the x is x³