This is differential calculus.I am doing the quotient rule: dy/dx = v(du/dx) - u(dv/dx) over v²
Question: If y = 1/x+1 , find dy/dx
Hence find the two values of x for which dy/dx = -1
Question: If y = 1/x+1 , find dy/dx
Hence find the two values of x for which dy/dx = -1
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You shouldn't need to use the quotient rule, as there is no function of x in the numerator. Chain rule is all you need:
y = (x+1)^(-1)
dy/dx = -(x+1)^(-2)
That is your answer. I'm not sure why you want to find when dy/dx = -1, as there is nothing in the original problem that states this. But if you need to do so, then:
-1 = -(x+1)^(-2)
(x+1)² = 1
x² + 2x + 1 = 1
x² + 2x = 0
x(x + 2) = 0
x = 0, -2
y = (x+1)^(-1)
dy/dx = -(x+1)^(-2)
That is your answer. I'm not sure why you want to find when dy/dx = -1, as there is nothing in the original problem that states this. But if you need to do so, then:
-1 = -(x+1)^(-2)
(x+1)² = 1
x² + 2x + 1 = 1
x² + 2x = 0
x(x + 2) = 0
x = 0, -2