How do you find the derivative of $y = \sqrt{x + 1}$ $y =sqrt(x+1)$ ?

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Answer 1 · 2014-09-21T16:49:21

By rewriting the square-root as the 1/2-power,

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

$y'=1/2(x+1)^{1/2-1}cdot(x-1)'$

$=1/2(x+1)^{-1/2}cdot 1$

$=1/{2sqrt{x+1]}$

How do you find the derivative of y =sqrt(x+1)y=√x+1y =sqrt(x+1)?