Express $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ in terms of x and y.

Find the value of $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ at the point on C where y = 1 and $x > 0$.

attempt at implicit differentiation     M1

EITHER

$\frac{{2x}}{y} - \frac{{{x^2}}}{{{y^2}}}\frac{{{\text{d}}y}}{{{\text{d}}x}} - 2 = \frac{1}{y}\frac{{{\text{d}}y}}{{{\text{d}}x}}$     A1A1

Note: Award A1 for each side.

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{\frac{{2x}}{y} - 2}}{{\frac{1}{y} + \frac{{{x^2}}}{{{y^2}}}}}{\text{ }}\left( { = \frac{{2xy - 2{y^2}}}{{{x^2} + y}}} \right)$     A1

OR

after multiplication by y

$2x - 2y - 2x\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}x}}\ln y + y\frac{1}{y}\frac{{{\text{d}}y}}{{{\text{d}}x}}$     A1A1

Note: Award A1 for each side.

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{2(x - y)}}{{1 + 2x + \ln y}}$     A1

[4 marks]

for $y = 1,{\text{ }}{x^2} - 2x = 0$

$x = (0{\text{ or) 2}}$     A1

for $x = 2$, $\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{2}{5}$     A1

[2 marks]

Most candidates were familiar with the concept of implicit differentiation and the majority found the correct derivative function. In part (b), a significant number of candidates didn’t realise that the value of x was required.

Most candidates were familiar with the concept of implicit differentiation and the majority found the correct derivative function. In part (b), a significant number of candidates didn’t realise that the value of x was required.