Q.Find $\dfrac{dy}{dx}$ of the function $x^y + y^x = 1$. OR If $y = 3\cos(\log x) + 4\sin(\log x)$, show that : $x^2 y_2 + x y_1 + y = 0$

This question offers a choice. Alternative 1 differentiates the implicit equation $x^y+y^x=1$ using logarithmic differentiation. Alternative 2 shows $y=3\cos(\log x)+4\sin(\log x)$ satisfies the given second-order relation.

Alternative 1 — find $\dfrac{dy}{dx}$ if $x^y+y^x=1$:

Let $u=x^y$ and $v=y^x$, so $u+v=1$.

Differentiate $u=x^y$ using logarithmic differentiation: $\ln u=y\ln x$. Differentiating both sides w.r.t. $x$ (using the product rule on the right, since $y$ is a function of $x$):

$\frac{1}{u}\frac{du}{dx}=\frac{dy}{dx}\ln x+y\cdot\frac1x$

$\frac{du}{dx}=x^y\left(\ln x\,\frac{dy}{dx}+\frac{y}{x}\right)$

Differentiate $v=y^x$ similarly: $\ln v=x\ln y$.

$\frac{1}{v}\frac{dv}{dx}=\ln y+x\cdot\frac1y\frac{dy}{dx}$

$\frac{dv}{dx}=y^x\left(\ln y+\frac{x}{y}\frac{dy}{dx}\right)$

Since $u+v=1$ is constant, $\dfrac{du}{dx}+\dfrac{dv}{dx}=0$:

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

Collect the $\dfrac{dy}{dx}$ terms:

$\frac{dy}{dx}\left(x^y\ln x+\frac{x\,y^x}{y}\right)=-\left(\frac{y\,x^y}{x}+y^x\ln y\right)$

$\frac{dy}{dx}=-\frac{y\,x^{y-1}+y^x\ln y}{x^y\ln x+x\,y^{x-1}}$

Alternative 2 — show $x^2y_2+xy_1+y=0$ for $y=3\cos(\log x)+4\sin(\log x)$:

First derivative (chain rule, $\dfrac{d}{dx}\log x=\dfrac1x$): …