A curve has equation $\arctan {x^2} + \arctan {y^2} = \frac{\pi }{4}$.

(a)     Find $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ in terms of x and y.

(b)     Find the gradient of the curve at the point where $x = \frac{1}{{\sqrt 2 }}$ and $y < 0$.

(a)     METHOD 1

$\frac{{2x}}{{1 + {x^4}}} + \frac{{2y}}{{1 + {y^4}}}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$     M1A1A1

Note:     Award M1 for implicit differentiation, A1 for LHS and A1 for RHS.

$\frac{{{\text{d}}y}}{{{\text{d}}x}} =  - \frac{{x\left( {1 + {y^4}} \right)}}{{y\left( {1 + {x^4}} \right)}}$     A1

METHOD 2

${y^2} = \tan \left( {\frac{\pi }{4} - \arctan {x^2}} \right)$

$= \frac{{\tan \frac{\pi }{4} - \tan \left( {\arctan {x^2}} \right)}}{{1 + \left( {\tan \frac{\pi }{4}} \right)\left( {\tan \left( {\arctan {x^2}} \right)} \right)}}$     (M1)

$= \frac{{1 - {x^2}}}{{1 + {x^2}}}$     A1

$2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ - 2x\left( {1 + {x^2}} \right) - 2x\left( {1 - {x^2}} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}}$     M1

$2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ - 4x}}{{{{\left( {1 + {x^2}} \right)}^2}}}$

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

$\left( { = \frac{{2x\sqrt {1 + {x^2}} }}{{\sqrt {1 - {x^2}} {{\left( {1 + {x^2}} \right)}^2}}}} \right)$

[4 marks]

(b)     ${y^2} = \tan \left( {\frac{\pi }{4} - \arctan \frac{1}{2}} \right)$     (M1)

$= \frac{{\tan \frac{\pi }{4} - \tan \left( {\arctan \frac{1}{2}} \right)}}{{1 + \left( {\tan \frac{\pi }{4}} \right)\left( {\tan \left( {\arctan \frac{1}{2}} \right)} \right)}}$     (M1)

Note:     The two M1s may be awarded for working in part (a).

$= \frac{{1 - \frac{1}{2}}}{{1 + \frac{1}{2}}} = \frac{1}{3}$     A1

$y =  - \frac{1}{{\sqrt 3 }}$     A1

substitution into $\frac{{{\text{d}}y}}{{{\text{d}}x}}$

$= \frac{{4\sqrt 6 }}{9}$     A1

Note: Accept $\frac{{8\sqrt 3 }}{{9\sqrt 2 }}$ etc.

[5 marks]

Total [9 marks]