Find the differential of (cos2x)^2

This is a question that can be answered in two ways: using the product rule and the rule for differentiating powers, we will do both to see how the two rules are equivalent. chain rule: split the function into its two parts, multiply the differential of each part by the other, undifferentiated part, and add the two results (easier to explain with whiteboard) (cos2x)^2 = cos2x * cos2x = 2(-sin2x)cos2x + 2(-2sin2x)cos2x = -4sin2xcos2x Powers rule: reduce the power, multiply by the old power and then multiply by the differential of that inside the power bracket. (cos2x)^2 = 2 * (cos2x)^1 * -2sin2x = -4sin2xcos2x

Unsurprisingly these two give the same answers, but can you see how using the product rule leads us to the powers rule shortcut? I would ask the student to try and prove this, and if successful to prove this for all powers for extra familiarity. (The further proof will be one by induction which I believe is in Further Maths in most syllabi so if the pupil is also taking this then more focus will be put onto it.) (proof) d/dx(f(x)^2) = f(x)f'(x) + f'(x)f(x) = 2f(x)f'(x) d/dx(f(x)^n) = (f(x)^n-1)f'(x) + d/dx(f(x)^n)f(x) = (f(x)^n-1)f'(x) + n-1(f(x)^n-1)f'(x) = nf(x)^n-1f'(x) which is the full power rule, we know it to be true for n = 2 and inductively for all higher integers so we have proved the power rule from the product rule. (If the student is advanced then I would go on to explain the way in which trigonemetric identities can be actually be thought of as infinite polynomials - this seems complex at first but actually can serve to demystify the use of trig functions which students feel less comfortable with that polynomials but function the same way. Similar methods can help the understanding of the exponential function.)