Q.Find $\dfrac{dy}{dx}$, if $y^x + x^y + x^x = a^b$. OR If $y = 3\cos(\log x) + 4\sin(\log x)$, show that $x^2 y_2 + x y_1 + y = 0$.

Take logarithms term-by-term and use implicit + logarithmic differentiation to isolate $dy/dx$; the OR part differentiates the given $y$ twice and substitutes to verify the stated second-order relation.

Part 1 — Find $\dfrac{dy}{dx}$ if $y^x+x^y+x^x=a^b$

Since $a^b$ is a constant, its derivative is $0$. Let $u=y^x,\ v=x^y,\ w=x^x$, so $u+v+w=a^b$.

For $u=y^x$: $\ln u = x\ln y \Rightarrow \dfrac{1}{u}\dfrac{du}{dx}=\ln y + \dfrac{x}{y}\dfrac{dy}{dx}$, so $\dfrac{du}{dx}=y^x\ln y + x\,y^{x-1}\dfrac{dy}{dx}$.

For $v=x^y$: $\ln v = y\ln x \Rightarrow \dfrac{1}{v}\dfrac{dv}{dx}=\dfrac{dy}{dx}\ln x + \dfrac{y}{x}$, so $\dfrac{dv}{dx}=x^y\ln x\,\dfrac{dy}{dx} + y\,x^{y-1}$.

For $w=x^x$: $\ln w = x\ln x \Rightarrow \dfrac{1}{w}\dfrac{dw}{dx}=\ln x+1$, so $\dfrac{dw}{dx}=x^x(1+\ln x)$.

Adding and setting the sum to zero:

$y^x\ln y + x\,y^{x-1}\dfrac{dy}{dx} + x^y\ln x\,\dfrac{dy}{dx} + y\,x^{y-1} + x^x(1+\ln x) = 0$

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

$\dfrac{dy}{dx}\big[x\,y^{x-1} + x^y\ln x\big] = -\big[y^x\ln y + y\,x^{y-1} + x^x(1+\ln x)\big]$

$\dfrac{dy}{dx} = -\dfrac{y^x\ln y + y\,x^{y-1} + x^x(1+\ln x)}{x\,y^{x-1} + x^y\ln x}$

OR — Part 2: If $y=3\cos(\log x)+4\sin(\log x)$, show $x^2y_2+xy_1+y=0$

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