Dany612
this chain rule problem is a bit confusing. x(2x3)^1/2



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giggles123
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3(x1)/(2x3)^.5

myininaya
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\[[x(2x3)^\frac{1}{2}]'= \text{ (apply product rule ) }[x]'(2x3)^\frac{1}{2}+x[(2x3)^\frac{1}{2}]'\]

myininaya
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\[[x]'=1\]
\[[(2x3)^\frac{1}{2}]'=\frac{1}{2}(2x3)^{\frac{1}{2}1}(2)\]

myininaya
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\[[x(2x3)^\frac{1}{2}]'=(2x3)^\frac{1}{2}+x(2x3)^\frac{1}{2}\]

myininaya
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\[(2x3)^\frac{1}{2}+\frac{x}{(2x3)^\frac{1}{2}}\]

myininaya
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\[(2x3)^\frac{1}{2} \cdot \frac{(2x3)^\frac{1}{2}}{(2x3)^\frac{1}{2}}+\frac{x}{(2x3)^\frac{1}{2}}\]

myininaya
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\[\frac{(2x3)+x}{(2x3)^\frac{1}{2}}\]

myininaya
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\[\frac{3x3}{\sqrt{2x3}} \]

myininaya
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\[\frac{3(x1)}{\sqrt{2x3}}\]