Find the derivative of the following function: 1/ax^(2) + bx+c ?

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gsenviro | College Teacher | (Level 1) Educator Emeritus

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The given function is f(x) = 1/(ax^2 +bx +c)

its derivative will be calculated as, f'(x) = d/dx [1/(ax^2 + bx+c)]

using the quotient rule of derivative, if f(x) = g(x)/h(x), then f'(x) = [g'(x)h(x)-g(x)h'(x)]/(h(x) x h(x))

We get, 

f'(x) = {[d/dx (a) x (ax^2 + bx+c)] - 1 x d/dx(ax^2 + bx+c) } / (ax^2 +bx+c)^2

      = -1 x (2ax + b)/(ax^2 +bx+c)^2

thus, f'(x) = -(2ax+b) / (ax^2 +bx+c)^2

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