Show that \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{{\cos }^2}x - {x^2}\sin x}}{{{{(x + \cos x)}^2}}},{\text{ }}x \geqslant 0\).

Find the equation of the tangent to C at the point \(\left( {\frac{\pi }{2},0} \right)\).

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{(x + \cos x)(\cos x - x\sin x) - x\cos x(1 - \sin x)}}{{{{(x + \cos x)}^2}}}\)     M1A1A1

Note: Award M1 for attempt at differentiation of a quotient and a product condoning sign errors in the quotient formula and the trig differentiations, A1 for correct derivative of “u”, A1 for correct derivative of “v”.

\( = \frac{{x\cos x + {{\cos }^2}x - {x^2}\sin x - x\cos x\sin x - x\cos x + x\cos x\sin x}}{{{{(x + \cos x)}^2}}}\)     A1

\( = \frac{{{{\cos }^2}x - {x^2}\sin x}}{{{{(x + \cos x)}^2}}}\)     AG

[4 marks]

the derivative has value –1     (A1)

the equation of the tangent line is \((y - 0) = ( - 1)\left( {x - \frac{\pi }{2}} \right)\left( {y = \frac{\pi }{2} - x} \right)\)     M1A1

[3 marks]

The majority of candidates earned significant marks on this question. The product rule and the quotient rule were usually correctly applied, but a few candidates made an error in differentiating the denominator, obtaining \( - \sin x\) rather than \(1 - \sin x\). A disappointing number of candidates failed to calculate the correct gradient at the specified point.

The majority of candidates earned significant marks on this question. The product rule and the quotient rule were usually correctly applied, but a few candidates made an error in differentiating the denominator, obtaining \( - \sin x\) rather than \(1 - \sin x\). A disappointing number of candidates failed to calculate the correct gradient at the specified point.