anonymous
  • anonymous
Find the derivative of g(x)= -3/ squareroot x
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
\[g(x)=\frac{-3}{\sqrt{x}}=\frac{-3}{x^{\frac{1}{2}}}=-3x^{\frac{-1}{2}}\]Now you can just use the power rule.
anonymous
  • anonymous
\[\frac{d}{dx}x^{n}=n\times x^{n-1}\]
anonymous
  • anonymous
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anonymous
  • anonymous
for petes sake to not use the product rule with constants
anonymous
  • anonymous
quotient rule then ?
anonymous
  • anonymous
no power rule
anonymous
  • anonymous
@Aylin wrote the method above
anonymous
  • anonymous
Power rule, Aylin have posted it up here.
anonymous
  • anonymous
\[y'= -\frac{ 3 }{ 2 }x^{-\frac{ 1 }{ 2 }}\]
anonymous
  • anonymous
like that @satellite73 @Aylin
anonymous
  • anonymous
You didn't execute the power rule properly.
anonymous
  • anonymous
\[g(x) = -3x ^{\frac{ -1 }{ 2 }}\]g'(x) will be: \[g'(x)=(-3*-\frac{ 1 }{ 2 })*x ^{-\frac{ 1 }{ 2 }-1}=\frac{ 3 }{ 2 } *x^{\frac{ -3 }{ 2 }} \iff g'(x)=\frac{ 3 }{ 2\sqrt{x^3} }\]
anonymous
  • anonymous
Okay but the thing is you use the power rule first and then take the derivative ?:S
anonymous
  • anonymous
No, the power rule is a way to differentiate. :]
anonymous
  • anonymous
@zairhenrique then why is it done twice here ?
anonymous
  • anonymous
It's only done once
anonymous
  • anonymous
I am really confused
anonymous
  • anonymous
the "power rule" is not the same as "write in exponential form" they are two different things
anonymous
  • anonymous
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 \[f'(x)=-3\times 2x^{-3-1}=-6x^{-4}\]
anonymous
  • anonymous
\[g(x) = \frac{ -3 }{ \sqrt{x}} \]
anonymous
  • anonymous
that is not "using the power rule twice" that is "write in exponential form" then "use the power rule"
anonymous
  • anonymous
\[g(x)=-3x^{\frac{ 1 }{ 2 }}\]
anonymous
  • anonymous
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
anonymous
  • anonymous
Yes i know
anonymous
  • anonymous
btw \[\frac{3}{\sqrt{x}}\neq -3x^{\frac{1}{2}}\]it is \[-3x^{-\frac{1}{2}}\]
anonymous
  • anonymous
ohh i see the problem make sure to write it in exponential form correctly
anonymous
  • anonymous
\[g'(x)=\frac{ -3 }{ 2 }x^\frac{ 1 }{ 2 }\]
anonymous
  • anonymous
ops that exponent is supposed ot be negative
anonymous
  • anonymous
\[= \frac{ -3 }{ 2\sqrt{x} }\]
anonymous
  • anonymous
It's not right... The -3/2 became positive and the x is x³
anonymous
  • anonymous
|dw:1370921862231:dw|
anonymous
  • anonymous
This is what i'm doing
anonymous
  • anonymous
g(x) = -3 * x^-1/2, not -3*x^1/2
anonymous
  • anonymous
OH
anonymous
  • anonymous
Is that why Im messing up
anonymous
  • anonymous
What's -3 * 1/2?
anonymous
  • anonymous
3/2
anonymous
  • anonymous
You sure?
anonymous
  • anonymous
what is half of -3?
anonymous
  • anonymous
- 3/2 **
anonymous
  • anonymous
OK, so that will be the coefficient :) Now what's 1/2 - 1?
anonymous
  • anonymous
-1/2
anonymous
  • anonymous
OK, so that will be the new exponent. Put it all together and you got the derivative! :) (-3/2)x^(-1/2)
anonymous
  • anonymous
|dw:1370922592191:dw|
anonymous
  • anonymous
Thank youuuu ! Finally understand lol
anonymous
  • anonymous
@FutureMathProfessor , the derivate is incorrect. The derivate of g(x) = -3/sqrtx is g'(x) = 3/2*sqrtx^3
anonymous
  • anonymous
\[g(x)=\frac{ -3 }{ \sqrt{x} }=-3*x ^{-\frac{ 1 }{ 2 }}=-3x ^{-\frac{ 1 }{ 2 }}\]Now, using the power rule, multiply the coefficient by -1/2 and subtract 1 from -1/2. So:\[g'(x)=-3*(-\frac{ 1 }{ 2 })*x ^{-\frac{ 1 }{ 2 }-1} = \frac{ 3 }{ 2 }*x ^{-\frac{ 3 }{2 }}=\frac{ 3 }{ 2\sqrt{x^3} }\]

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