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zysi [14]
3 years ago
10

given ghi with g(4 -3), H (-4, 2), and I (2, 4), find the perpendicular bisector of HI in standard form

Mathematics
1 answer:
vredina [299]3 years ago
7 0
GI is the bisector of the perpendicular bisector.

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Answer:

\dfrac{dy}{dx}=- \dfrac{1}{\left(x - 1\right) \left(x + 1\right)}

Step-by-step explanation:

Given:

y = \ln{\left(\dfrac{x + 1}{x - 1}\right)^{\frac{1}{2}}

using the properties of log we can take the power 1/2 and multiply it.

y = \dfrac{1}{2}\ln{\left(\dfrac{x + 1}{x - 1}\right)

now we can differentiate:

\dfrac{dy}{dx} = \dfrac{1}{2}\dfrac{1}{\left(\dfrac{x + 1}{x - 1}\right)}\left(\dfrac{d}{dx}\left(\dfrac{x+1}{x-1}\right)\right)

\dfrac{dy}{dx} = \dfrac{1}{2}\left(\dfrac{x - 1}{x + 1}\right)\left(- \dfrac{2}{\left(x - 1\right)^{2}}\right)

\dfrac{dy}{dx}=- \dfrac{1}{\left(x - 1\right) \left(x + 1\right)}

this is our answer!

8 0
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Answer:

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Step-by-step explanation:

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