anonymous
  • anonymous
f(x)=sqrt(3x+1), find the derivative at a
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
You can f(x) as : \[(3x+1)^{\frac{ 1 }{ 2 }}\]Still the same meaning
anonymous
  • anonymous
there's a short cut but I'll show you the proper one \[\lim_{\Delta x \rightarrow 0 } \frac{ f(x+\Delta x)-f(x) }{ \Delta x }\]
anonymous
  • anonymous
\[\frac{ \sqrt{3(x+\Delta x) + 1} -\sqrt{3x+1} }{ \Delta x }\]

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KenLJW
  • KenLJW
f(x)=sqrt(3x+1) f'(x)=[(1/2)/sqrt(3x+1)] 3 chain rule
anonymous
  • anonymous
Multiply by the conjugate
anonymous
  • anonymous
\[\frac{ [3(x+\Delta x) +1] - 3x+1 }{ \Delta x (\sqrt{3(x+\Delta x)+1}+\sqrt{3x+1)} }\]
anonymous
  • anonymous
\[ 3x + 3 \Delta x +1 -3x -1 \] on the numerator and same thing on the denominator
anonymous
  • anonymous
also, my bad, it's -3x -1
anonymous
  • anonymous
after you cancel out everything, 3 Delta x remains, you simplify and cancel out delta x, and you'll left out with \[\frac{ 3 }{ \sqrt{3x+3\Delta x +1}+\sqrt{3x+1} }\]
anonymous
  • anonymous
and since Delta x is approaching 0, you plug in 0 and you'll get 3/2sqrt(3x+1)
KenLJW
  • KenLJW
Nice development muskie, good to know procedure but why reinvent the wheel.
anonymous
  • anonymous
I just went over the basic steps. and Thank you for your compliment.
anonymous
  • anonymous
There's a shortcut of course.

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