If ${\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^n=2$, the $\frac{dy}{dx}$at $(a,b)$ is

${\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^n=2$

Differentiate with respect to $\text{'}x\text{'}$

$n{\left(\frac{x}{a}\right)}^{n-1}.\left(\frac{1}{a}\right)+n.{\left(\frac{y}{b}\right)}^{n-1}.\frac{y\text{'}}{b}=0$

$n.{\left[\frac{y}{b}\right]}^{n-1}.\frac{y\text{'}}{b}=-n.{\left(\frac{x}{a}\right)}^{n-1}.\left(\frac{1}{a}\right)$

$\text{at\hspace{0.17em}\hspace{0.17em}}(a,b)$

${\left(\frac{b}{b}\right)}^{n-1}.\frac{y\text{'}}{b}=-{\left(\frac{a}{a}\right)}^{n-1}.\left(\frac{1}{a}\right)$

$\frac{y\text{'}}{b}=-\frac{1}{a}$

$y\text{'}=-\frac{b}{a}$