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6.1 Introductory Calculus • 229Therefore, you must use the limit definition of the derivative.> fx(0,0) := limit( ( f(h,0) - f(0,0) )/h, h=0 );fx(0, 0) := 0At (x, y) = (r cos(θ), r sin(θ)) the value of fx is> fx( r*cos(theta), r*sin(theta) );r sin(θ) (r 2 cos(θ) 2 − r 2 sin(θ) 2 ) r 3 cos(θ) 2 sin(θ)r 2 cos(θ) 2 + r 2 sin(θ) 2 + 2r 2 cos(θ) 2 + r 2 sin(θ) 2− 2 r3 cos(θ) 2 sin(θ) (r 2 cos(θ) 2 − r 2 sin(θ) 2 )(r 2 cos(θ) 2 + r 2 sin(θ) 2 ) 2> combine( % );34 r sin(3 θ) − 1 r sin(5 θ)4As the distance r from (x, y) to (0, 0) tends to zero, so does |fx(x, y)−fx(0, 0)|.> Limit( abs( % - fx(0,0) ), r=0 );lim3r→0 ∣4 r sin(3 θ) − 1 ∣ ∣∣∣4 r sin(5 θ)> value( % );0Hence, fx is continuous at (0, 0).By symmetry, the same arguments apply to the derivative of f withrespect to its second parameter, y.> fy := D[2](f);fy := (x, y) → x (x2 − y 2 ) x y 2x 2 + y 2 − 2x 2 + y 2 − 2 x y2 (x 2 − y 2 )(x 2 + y 2 ) 2