Question

Maths 6 marks question.... Continuity and differentiability

Answer 1

Now,

$x^{y} + y^{x} = 1 \implies e^{y\log(x)} + e^{x\log(y)} = 1.$

Differentiating implicitly with respect to x gives:

$e^{y\log(x)}\left(\log(x)\frac{dy}{dx} + \frac{y}{x}\right) + e^{x\log(y)}\left(\log(y) + \frac{y}{x}\frac{dy}{dx} \right) = 0$

$\implies x^{y}\log(x)\frac{dy}{dx} + yx^{y-1} + y^{x}\log(y) + xy^{x-1}\frac{dy}{dx} = 0$

$\implies (x^{y}\log(x) + xy^{x-1})\frac{dy}{dx} = -yx^{y-1} - y^{x}\log(y)$

$\implies \frac{dy}{dx} = \underline{\underline{\frac{-yx^{y-1} - y^{x}\log(y)}{x^{y}\log(x) + xy^{x-1}}}}.$