Using the first principles, find the derivative of $y = \frac{3 - x}{\sqrt{2 - x}}$ $y=(3-x)/sqrt(2-x)$ at x=-2?

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Answer 1 · 2018-06-19T19:42:45

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

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

$y' = (-1) (2 - x)^(-1/2) + (3 - x) (-1/2)(2 - x)^(-3/2)(-1)$

$y' = - 1 / (2 - x)^(1/2) + (3 - x) / (2 * (2 - x)^(3/2))$

$y' = (-2* (2 - x)+ 3 - x) / (2 * (2 - x)* (2 - x)^(1/2))$

$y' = (x - 1) / (2 * (2 - x)* sqrt(2 - x))$

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