Question

Find dy/dx tanx)y=tany)x

Answer 1

you mean, we have to find dy/dx of equation, tan(x/y) = tan(y/x) , right ?

tan(x/y) = tan(y/x)

differentiating both sides with respect to x,

or, d[tan(x/y)]/dx = d[tan(y/x)]/dx

or, sec²(x/y) × d(x/y)/dx = sec²(y/x) × d(y/x)

or, sec²(x/y) × [(y × dx/dx - x × dy/dx)/y²] = sec²(y/x) × [ (x × dy/dx - y × dx/dx )/x²]

or, sec²(x/y) × (y - xy')/y² =sec²(y/x) × (xy' - y)/x²

or, x²sec²(x/y) × (y - xy') = y²sec²(y/x) × (xy' - y)

or, x²ysec²(x/y) + y³ sec²(y/x) = y²xy' sec²(y/x) + x³y'sec²(x/y)

or, y' = [x²ysec²(x/y) + y³sec²(y/x)]/[y²xsec²(y/x) + x³sec²(x/y) ]

hence, differentiation of given equation is dy/dx = [x²ysec²(x/y) + y³sec²(y/x)]/[y²xsec²(y/x) + x³sec²(x/y) ]