What is the derivative of $y = {[\frac{x - 2}{k}]}^{\frac{1}{2}}$ $y = [(x-2)/k]^(1/2)$ where k is a constant?

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Answer 1 · 2017-10-20T02:10:58

Look below.

where $k \neq 0$ $k!=0$

$\frac{d y}{d x} = \frac{1}{2} {[\frac{x - 2}{k}]}^{- \frac{1}{2}} \cdot \frac{d}{d x} (\frac{x - 2}{k})$ $dy/dx=1/2[(x-2)/k]^(-1/2)*d/dx((x-2)/k)$

$= \frac{1}{2} {[\frac{x - 2}{k}]}^{- \frac{1}{2}} \cdot (\frac{1}{k})$ $=1/2[(x-2)/k]^(-1/2)*(1/k)$

$= \frac{1}{2 \sqrt{k (x - 2)}}$ $=1/(2sqrt(k(x-2))$

What is the derivative of y = [(x-2)/k]^(1/2)y=[x−2k]12y = [(x-2)/k]^(1/2) where k is a constant?