y= [(x-3)/(x^2+1)]^2
I know that I would use the product rule on the whole equation inside the brackets. But would i also do the quotient rule bc of the division?
Correction**** I would use the chain rule. not product rule
your "first line derivative" would be
dy/dx = 2[(x-3)/(x^2 + 1) ] [ (x^2+1)(1) - (x-3)(2x)]/(x^2 + 1)^2
In the given equation, y is defined as [(x-3)/(x^2+1)]^2. To find the derivative of y, we need to use the chain rule since the function inside the brackets is raised to the power of 2. The chain rule states that if we have a composition of functions, f(g(x)), the derivative is given by f'(g(x)) * g'(x).
Let's break down the function: f(g(x)) = g(x)^2, where g(x) = (x-3)/(x^2+1).
First, we can find the derivative of g(x) using the quotient rule:
g'(x) = ([(x^2+1) * (1)] - [(x-3) * (2x)]) / (x^2+1)^2
= (x^2 + 1 - 2x^2 + 6x) / (x^2 + 1)^2
= (-x^2 + 6x + 1) / (x^2 + 1)^2
Next, we can find the derivative of f(g(x)) using the chain rule:
f'(g(x)) = 2 * g(x)^(2-1) * g'(x)
= 2 * g(x) * g'(x)
= 2 * [(x-3)/(x^2+1)] * [(-x^2 + 6x + 1) / (x^2 + 1)^2]
So, the derivative of y is given by:
y' = 2 * [(x-3)/(x^2+1)] * [(-x^2 + 6x + 1) / (x^2 + 1)^2]
Therefore, we only need to use the chain rule to find the derivative and not the quotient rule since there is no division occurring inside the brackets.