Question

$x^{a} . y^{b} = (x+y)^{a+b}$
show that $\frac{dy}{dx} = \frac{y}{x}$

Answer 1

$x^a.y^b=(x+y)^{a+b},\ \ \ \ \ \ Differentiating\ wrt\ x,\\ \\a.x^{a-1}.y^b+x^a.b.y^{b-1}.y'=(a+b)(x+y)^{a+b-1}(1+y')\\ \\(ay+bx.y')\ x^{a-1}.y^{b-1}=(a+b)(x+y)^{a+b}\frac{1}{x+y}(1+y')\\ \\(ay+bx.y')\ x^{a-1}.y^{b-1}=(a+b)(x^ay^b)\frac{1}{x+y}(1+y')\\ \\(ay+bx.y')=(a+b)\frac{xy}{x+y}(1+y')\\ \\ay-\frac{(a+b)xy}{x+y}=y'.[\frac{(a+b)xy}{x+y}-bx]\\ \\\frac{axy+ay^2-axy-bxy}{x+y}=y'.\frac{axy+bxy-bx^2-bxy}{x+y}\\ \\y(ay-bx)=y'(ay-bx)x\\ \\y'=\frac{y}{x}$