anonymous
  • anonymous
Derivative Question: Help (attached below)
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
NoelGreco
  • NoelGreco
Get rid of the radicals.
anonymous
  • anonymous
hint:\[ x^5\sqrt{x}=x^{5+\frac{1}{2}}=x^{\frac{11}{2}}\]

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anonymous
  • anonymous
right.. 3x^(11/2)
NoelGreco
  • NoelGreco
That's it, and go with a negative exponent on the second term to avoid the quotient rule.
anonymous
  • anonymous
+ -4*x^-7/2 ?
NoelGreco
  • NoelGreco
Yes.
anonymous
  • anonymous
i got: 3x^(11/2)+14/x^9/2..
anonymous
  • anonymous
NoelGreco
  • NoelGreco
\[3x ^{\frac{ 11 }{ 2 }}-4x ^{\frac{ -7 }{ 2 }}\] Now take the derivative.
anonymous
  • anonymous
14x/x^9/2
anonymous
  • anonymous
?
NoelGreco
  • NoelGreco
The derivative of the second term is\[\frac{ d }{ dx }3x ^{\frac{ 11 }{ 2 }}=\frac{ 33 }{ 2 }x ^{\frac{ 9 }{ 2 }}\] I don't know how you're trying to take the derivative, but you simply use the power rule on each term.
anonymous
  • anonymous
huh?
anonymous
  • anonymous
from 3x^ (11/2) + 14x^(-9/2)
anonymous
  • anonymous
doesn't the x9/2 down
anonymous
  • anonymous
@NoelGreco hey sorry to interrupt. but i just needed to figure this out before leaving..
anonymous
  • anonymous
\[\frac{d}{dx}14x^{-\frac{9}{2}}=-\frac{9}{2}\times 14x^{-\frac{9}{2}-1}\]whatever that is
anonymous
  • anonymous
this is my work so far:
1 Attachment
anonymous
  • anonymous
Before applying the power rule combine the terms as told you previously . \[\Large f(x)=3x^5.x^\frac{1}{2}-\frac{-4}{x^3.x^\frac{1}{2}}\] combining the terms \[\Large f(x)=3x^\frac{11}{2}-\frac{4}{x^\frac{7}{2}}\] \[\Large f(x)=3x^\frac{11}{2}-4x^{-\frac{7}{2}}\] now Apply the Power rule.
anonymous
  • anonymous
thank you are you still there @sami-21
anonymous
  • anonymous
i have one question if you dont mind
anonymous
  • anonymous
what if they ask: derivative of sqrt(6x)
anonymous
  • anonymous
i dont know why i always messed them up
anonymous
  • anonymous
i thought it would be like this: = 6x^(1/2) then =3x^(-1/2) =3/(x^(1/2)
anonymous
  • anonymous
anonymous
  • anonymous
yes that is correct.

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