Not really. For your second method, you have to include multiplication by dy/dx as well since you're using the chain rule i.e implicit differentiation

letting y = √x = x¹/²

d/dx (y(7y - 1))

= d/dx (7y² - y)

= (14y - 1) dy/dx

= (14y - 1) (½x-¹/²)

= (14√x - 1)(1 / 2√x)

= 7 - 1 / 2√x

Thanks for the spot! I'd forgot about applying chain rule using dy/dx here. Yeap, your method is correct.

However, since implicit differentiation is not taught at O levels, the question should be looking for :

① using product rule directly

and

② expanding the expression first , then differentiate.

① would be :


d/dx (√x (7√x - 1))


= (d/dx √x) (7√x - 1) + √x (d/dx (7√x - 1))

= (1 / 2√x) (7√x - 1) + √x (7 / 2√x - 0)

= 7/2 - 1 / 2√x + 7/2

= 7 - 1 / 2√x

No problem