Question

Could someone please help and explain to me how to find the derivative of y=1/((3x)^3)

Answer 1

Quotient Rule: y' = [f'(x)g(x) - f(x)g'(x)]/g(x)^2
f = numerator, g = denominator

y' = [f'(x^2 - 3x - 1)g(x - 2) - f(x^2 - 3x - 1)g'(x - 2)]/g(x - 2)^2
y' = [(2x - 3)(x - 2) - (1)(x^2 - 3x - 1)]/(x - 2)^2
= (2x^2 - 7x + 6 - x^2 + 3x + 1)/(x - 2)^2
= (x^2 - 4x + 7)/(x - 2)^2

Answer 2

um sorry, but I've only learned up to chain rule in class, so if possible could you explain it that way?

Answer 3

simplify first, y = 1/((3x)^3) = 1/(27x^3) = 1/27 * x^-3
so, y ' = 1/27 * -3x^-4 = - 3/27 x^-4 = -1/(9x^4)

Answer 4

\[\frac{ \alpha y(x) }{ \alpha x } = \frac{ 1 }{ -9x^4 }\]

Answer 5

thanks guys :)