y = 4x/(x^2+5). a) Find dy/dx, writing your answer as a single fraction in its simplest form. b) Hence find the set of values of x for which dy/dx < 0

a) We need to differentiate this equation using the quotient rule (Given that it is a fraction with an x term on both the top and bottom of the fraction). We assign the numerator and denominator as follows: u = 4x, v = x^2+5, and so using the chain rule to differentiate with respect to x in both cases we get: u' = 4, v' = 2x, we now use the following quotient rule equation: dy/dx = (u'v-uv')/v^2, subbing in our values we have: 4(x^2+5)-(4x)(2x)/(x^2+5)^2 . Which is simply: 20-4x^2/(x^2+5)^2
b) To find the set of values of x, we first set the simplified form of dy/dx that we found in the previous part of the question = 0. And so we have that 20-4x^2/(x^2+5)^2 = 0. In order for this equality to be the case one of two things must be true; either the numerator is equal to zero OR the denominator is equal to infinity. The case that we can prove is when the numerator is equal to zero. And so setting 20-4x^2 = 0 we can see that x^2 = 20/4 = 5. However our question wanted us to find the set of values of x for which dy/dx < 0, not = 0. We therefore update this knowledge onto our value and so have x^2 > 5, rooting both sides gives x > +/- sqrt(5). After plotting a simple aid diagram to help us make sense of when this is true we find our set of values of x to be x < -sqrt(5), x > sqrt(5)