Asked by mo
Find y' :
y= (8-2x^2)/(root of (8-x^2))
y= (8-2x^2)/(root of (8-x^2))
Answers
Answered by
Reiny
using the product rule
dy/dx = 1/2(8-x^2)^(-1/2)(-2x) + (8-x^2)^(1/2)(-4x)
= -(8-x^2)^(-1/2) - 4x(8-x^2)(1/2)
= -(8-x^2)^(-1/2)[8-x^2 + 4x]
= (8-x^2)^(-1/2)(x^2 - 4x - 8)
dy/dx = 1/2(8-x^2)^(-1/2)(-2x) + (8-x^2)^(1/2)(-4x)
= -(8-x^2)^(-1/2) - 4x(8-x^2)(1/2)
= -(8-x^2)^(-1/2)[8-x^2 + 4x]
= (8-x^2)^(-1/2)(x^2 - 4x - 8)
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