Zi Xiang's answer to Ginella's Secondary 4 A Maths Singapore question.
since numerator is a constant with no x, then instead of quotient rule, may be easier to just use …
(d/dx)[a x^n] = (a)(n)x^(n-1)
thus …
(d/dx)[12 (3-2x)^(-3)]
=12(-3)(-2)(3-2x)^(-4)
=72/(3-2x)^4.
(d/dx)[a x^n] = (a)(n)x^(n-1)
thus …
(d/dx)[12 (3-2x)^(-3)]
=12(-3)(-2)(3-2x)^(-4)
=72/(3-2x)^4.
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