Question Number 104542 by mohammad17 last updated on 22/Jul/20

Answered by Dwaipayan Shikari last updated on 22/Jul/20

2)  tan^(−1) ((y/x))=log(x^2 +y^2 )  (((1/x).(dy/dx)−(y/x^2 ))/((y^2 /x^2 )+1))=((2x+2y(dy/dx))/(x^2 +y^2 ))  x(dy/dx)−y=2x+2y(dy/dx)  (dy/dx)=((2x+y)/(x−2y))

Commented bymohammad17 last updated on 22/Jul/20

thank you sir

Answered by Dwaipayan Shikari last updated on 22/Jul/20

xy=y^(tanx)   logx+logy=tanx logy  (1/x)+(1/y) (dy/dx)=tanx.(1/y) (dy/dx)+sec^2 xlogy  (1/y) (dy/dx)(1−tanx)=((xsec^2 logy−1)/x)  ((tanx)/(log(xy))).(dy/dx)=((xsec^2 xlogy−1)/(x(1−tanx)))  (dy/dx)=((log(xy))/(tanx))(((xsec^2 xlogy−1)/(x(1−tanx))))

Commented bymohammad17 last updated on 22/Jul/20

Commented bymohammad17 last updated on 22/Jul/20

sir is the solution true ?

Commented byDwaipayan Shikari last updated on 22/Jul/20

yes