Let \(h(x) = \frac{{6x}}{{\cos x}}\) . Find \(h'(0)\) .

METHOD 1 (quotient)

derivative of numerator is 6     (A1)

derivative of denominator is \( - \sin x\)     (A1)

attempt to substitute into quotient rule     (M1)

correct substitution     A1

e.g. \(\frac{{(\cos x)(6) - (6x)( - \sin x)}}{{{{(\cos x)}^2}}}\)

substituting \(x = 0\)     (A1)

e.g. \(\frac{{(\cos 0)(6) - (6 \times 0)( - \sin 0)}}{{{{(\cos 0)}^2}}}\)

\(h'(0) = 6\)     A1     N2

METHOD 2 (product)

\(h(x) = 6x \times {(\cos x)^{ - 1}}\)

derivative of 6x is 6     (A1)

derivative of \({(\cos x)^{ - 1}}\) is \(( - {(\cos x)^{ - 2}}( - \sin x))\)     (A1)

attempt to substitute into product rule     (M1)

correct substitution     A1

e.g. \((6x)( - {(\cos x)^{ - 2}}( - \sin x)) + (6){(\cos x)^{ - 1}}\)

substituting \(x = 0\)    (A1)

e.g. \((6 \times 0)( - {(\cos 0)^{ - 2}}( - \sin 0)) + (6){(\cos 0)^{ - 1}}\)

\(h'(0) = 6\)     A1     N2

[6 marks]

The majority of candidates were successful in using the quotient rule, and were able to earn most of the marks for this question. However, there were a large number of candidates who substituted correctly into the quotient rule, but then went on to make mistakes in simplifying this expression. These algebraic errors kept the candidates from earning the final mark for the correct answer. A few candidates tried to use the product rule to find the derivative, but they were generally not as successful as those who used the quotient rule. It was pleasing to note that most candidates did know the correct values for the sine and cosine of zero.