Find the gradient of the tangent to the curve ${x^3}{y^2} = \cos (\pi y)$ at the point (−1, 1) .

METHOD 1

$3{x^2}{y^2} + 2{x^3}y\frac{{{\text{d}}y}}{{{\text{d}}x}} = - \pi \sin (\pi y)\frac{{{\text{d}}y}}{{{\text{d}}x}}$     A1A1A1

At $( - 1,{\text{ }}1),{\text{ }}3 - 2\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$     M1A1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{3}{2}$     A1

[6 marks]

METHOD 2

$3{x^2}{y^2} + 2{x^3}y\frac{{{\text{d}}y}}{{{\text{d}}x}} = - \pi \sin (\pi y)\frac{{{\text{d}}y}}{{{\text{d}}x}}$     A1A1A1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{3{x^2}{y^2}}}{{ - \pi \sin (\pi y) - 2{x^3}y}}$     A1

At $( - 1,{\text{ }}1),{\text{ }}\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{3{{( - 1)}^2}{{(1)}^2}}}{{ - \pi \sin (\pi ) - 2{{( - 1)}^3}(1)}} = \frac{3}{2}$     M1A1

[6 marks]

A large number of candidates obtained full marks on this question. Some candidates missed $\pi$ and/or $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ when differentiating the trigonometric function. Some candidates attempted to rearrange before differentiating, and some made algebraic errors in rearranging.