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]

[N/A]