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Tanzania [10]
3 years ago
8

Find the derivative of |x|/(x-1)

Mathematics
1 answer:
ki77a [65]3 years ago
4 0
\bf \cfrac{|x|}{x-1}\iff \cfrac{\sqrt{x^2}}{x-1}\iff \cfrac{(x^2)^{\frac{1}{2}}}{x-1}
\\\\\\
\textit{using the quotient rule}
\\\\\\
\cfrac{dy}{dx}=\cfrac{\frac{1}{2}(x^2)^{-\frac{1}{2}}\cdot 2x(x-1)-(x^2)^{\frac{1}{2}}\cdot 1}{(x-1)^2}

\bf \cfrac{dy}{dx}=\cfrac{\frac{x(x-1)-(x^2)^{\frac{1}{2}}}{(x^2)^{\frac{1}{2}}}}{(x-1)^2}\implies \cfrac{dy}{dx}=\cfrac{x(x-1)-(x^2)^{\frac{1}{2}}}{(x^2)^{\frac{1}{2}}(x-1)^2}
\\\\\\\cfrac{dy}{dx}=\cfrac{x(x-1)-|x|}{|x|(x-1)^2}
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First we need to understand that moving upwards would be addition to y-coordinates, while moving downwards would be deduction to the y-coordinates.

Moving to rightwards would be addition to x-coordinates and leftwards would be deduction to x-coordinates.

As we moved down by 1 unit and up for 2 units, it means that we are originally upward for one unit and downward by 2 units, which makes downwards for 1 units and would be -1 to the y-coordinates:

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3 years ago
Task:
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3200 = 300(10.6_)

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