`x^2 + xy - y^2 = 4` Find `(dy/dx)` by implicit differentiation.
2 Answers
hkj1385 | (Level 1) Assistant Educator
Note:- 1) If y = x^n ; then dy/dx = n*x^(n-1) ; where 'n' = real number
2) If y = u*v ; where both u & v are functions of 'x' ; then
dy/dx = u*(dv/dx) + v*(du/dx)
3) If y = k ; where k = constant ; then dy/dx = 0
Now, the given function is :-
(x^2) + xy - (y^2) = 4
Differentiating both sides w.r.t 'x' we get,
2x + x*(dy/dx) + y - 2y*(dy/dx) = 0
or, (2x+y) = (2y-x)*(dy/dx)
or, dy/dx = (2x+y)/(2y-x)
balajia | College Teacher | (Level 1) eNoter
`x^2+xy-y^2=4`
Differentiating with respect to x. We get
`2x+(y+x(dy/dx))-2y(dy/dx)=0`
`(2x+y)+(x-2y)(dy/dx)=0`
`dy/dx=(2y-x)/(2x+y)`