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Dany612

  • 4 years ago

this chain rule problem is a bit confusing. x(2x-3)^1/2

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  1. giggles123
    • 4 years ago
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    3(x-1)/(2x-3)^-.5

  2. myininaya
    • 4 years ago
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    \[[x(2x-3)^\frac{1}{2}]'= \text{ (apply product rule ) }[x]'(2x-3)^\frac{1}{2}+x[(2x-3)^\frac{1}{2}]'\]

  3. myininaya
    • 4 years ago
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    \[[x]'=1\] \[[(2x-3)^\frac{1}{2}]'=\frac{1}{2}(2x-3)^{\frac{1}{2}-1}(2)\]

  4. myininaya
    • 4 years ago
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    \[[x(2x-3)^\frac{1}{2}]'=(2x-3)^\frac{1}{2}+x(2x-3)^\frac{-1}{2}\]

  5. myininaya
    • 4 years ago
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    \[(2x-3)^\frac{1}{2}+\frac{x}{(2x-3)^\frac{1}{2}}\]

  6. myininaya
    • 4 years ago
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    \[(2x-3)^\frac{1}{2} \cdot \frac{(2x-3)^\frac{1}{2}}{(2x-3)^\frac{1}{2}}+\frac{x}{(2x-3)^\frac{1}{2}}\]

  7. myininaya
    • 4 years ago
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    \[\frac{(2x-3)+x}{(2x-3)^\frac{1}{2}}\]

  8. myininaya
    • 4 years ago
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    \[\frac{3x-3}{\sqrt{2x-3}} \]

  9. myininaya
    • 4 years ago
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    \[\frac{3(x-1)}{\sqrt{2x-3}}\]

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