Y=3x^2+6x-7/(x+1)^2 Show that dy/dx = 20/(x+1)^3

1 answer:

3 0

Apply the quotient rule
d(uv) = v du/dx - u dv/dx
dx ----------------------
v^2

du/dx = 6x + 6
dv/dx = 2(x + 1)
v^2 = (x + 1)^4 so we have

dy dx = (x + 1)^2 .* (6x + 6) - (3x^2 + 6x - 7) * 2(x + 1) / (x + 1)^4

= (x + 1) [(6(x + 1)^2 - 2(3x^2 + 6x - 7) ] / (x + 1)^4

= (x^2 + 12x +6 - 6x^2 - 12x + 14 ) / (x + 1)^3

= 20 / (x + 1)^3 Answer

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