If y ^ x = xfind * (dy)/(dx)

1.	If y ^ x = xfind * (dy)/(dx)
Answer» $\large\underline{\sf{Given- }}$ $\rm :\longmapsto\: {y}^{x} = x$ On taking log on both sides, we get $\rm :\longmapsto\: log( {y}^{x} ) = log(x)$ $\rm :\longmapsto\:x log(y) = log(x)$ Now differnatiate both sides W. r. t. x, we get $\rm :\longmapsto\:\dfrac{d}{dx} (x \: logy) = \dfrac{d}{dx} logx$ We KNOW that, $\: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf{ \: \dfrac{d}{dx} u.v = v\dfrac{d}{dx} u + u\dfrac{d}{dx} v}}$ and $\: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf{ \: \dfrac{d}{dx} logx = \dfrac{1}{x} }}$ On APPLYING this, we get $\rm :\longmapsto\:x\dfrac{d}{dx} logy + logy\dfrac{d}{dx} x = \dfrac{1}{x}$ $\rm :\longmapsto\:x \times \dfrac{1}{y} \dfrac{dy}{dx} + logy \times 1 = \dfrac{1}{x}$ $\rm :\longmapsto\:\dfrac{x}{y}\dfrac{dy}{dx} = \dfrac{1}{x} - logy$ $\rm :\longmapsto\:\dfrac{x}{y}\dfrac{dy}{dx} = \dfrac{1 - xlogy}{x}$ $\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{y(1 - xlogy)}{ {x}^{2} }$ ADDITIONAL INFORMATION:- $\boxed{ \bf{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} }}$ $\boxed{ \bf{ \: \dfrac{d}{dx} x = 1}}$ $\boxed{ \bf{ \: \dfrac{d}{dx} k = 0}}$ $\boxed{ \bf{ \: \dfrac{d}{dx} sinx = cosx}}$ $\boxed{ \bf{ \: \dfrac{d}{dx} cosx = - sinx}}$ $\boxed{ \bf{ \: \dfrac{d}{dx} tanx = {sec}^{2} x}}$ $\boxed{ \bf{ \: \dfrac{d}{dx} secx = secxtanx}}$ $\boxed{ \bf{ \: \dfrac{d}{dx} {a}^{x} = {a}^{x} log(a)}}$

Answer»

$\large\underline{\sf{Given- }}$