If $x^{13}y^{7}=(x+y)^{20}$ , then $\frac{dy}{dx}$
directly doing it makes it very complicated so, I did this $\left(\frac{x}{y}\right)^{13}=\left(1+\frac{x}{y} \right)^{20}$. following are the options for solution (a) $\frac{y^2}{x^2}$ (b)$\frac{x^2}{y^2}$ (c)$\frac{x}{y}$ (d)$\frac{y}{x}$ thanks for any hints.
I differentiated directly,
$$13x^{12}y^7+7x^{13}y^6y'=20(x+y)^{19}(1+y')$$
$${13x^{13}y^7\over {x}}+{7x^{13}y^7y'\over y}={20(x+y)^{20}(1+y')\over {(x+y)}}$$
$$(x+y)(13y+7xy')=20xy(1+y')$$
Cancelling $x^{13}y^7$ and rearranging we get,
$$y'={13y^2-7xy\over {13xy-7x^2}}$$
Divide the numerator and the dinominator by$x^2$
$$y'={13({y\over x})^2-7({y\over x})\over {13({y\over x})-7}}$$
$$y'={y\over x}$$
