How do you differentiate $y = 2 / [3sqrt(x^2 - 5x)]$ ?

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Answer 1 · 2016-01-30T10:09:45

$(4x-10)/(3sqrt((x^2-5x)^3))$

Here,
$y=2/(3sqrt(x^2-5x))$

so,
$(dy)/(dx)$

$=d/(dx)(2/(3sqrt(x^2-5x)))$

$=2/3d/(dx)(1/(sqrt(x^2-5x)))$

$=2/3*1/((x^2-5x)^(3/2))d/(dx)(x^2-5x)$

$=2/3*1/((x^2-5x)^(3/2))(2x-5)$

$=(4x-10)/(3sqrt((x^2-5x)^3))$

How do you differentiate y = 2 / [3sqrt(x^2 - 5x)] y = 2 / [3sqrt(x^2 - 5x)] ?