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# Using the first principles, find the derivative of y=(3-x)/sqrt(2-x) at x=-2?

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?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Jun 19, 2018

there is no derivative at that point: infinity by one side, out of domain by the other side

#### Explanation:

$y = \left(3 - x\right) {\left(2 - x\right)}^{- \frac{1}{2}}$

$y ' = \left(- 1\right) {\left(2 - x\right)}^{- \frac{1}{2}} + \left(3 - x\right) \left(- \frac{1}{2}\right) {\left(2 - x\right)}^{- \frac{3}{2}} \left(- 1\right)$

$y ' = - \frac{1}{2 - x} ^ \left(\frac{1}{2}\right) + \frac{3 - x}{2 \cdot {\left(2 - x\right)}^{\frac{3}{2}}}$

$y ' = \frac{- 2 \cdot \left(2 - x\right) + 3 - x}{2 \cdot \left(2 - x\right) \cdot {\left(2 - x\right)}^{\frac{1}{2}}}$

$y ' = \frac{x - 1}{2 \cdot \left(2 - x\right) \cdot \sqrt{2 - x}}$

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