Question

If y ^ x = xfind * (dy)/(dx)​

Answer 1

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

$\rm :\longmapsto\: {y}^{x} = x$

• 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:-

$\boxed{ \bf{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} }}$

$\boxed{ \bf{ \: \dfrac{d}{dx} x = 1}}$

$\boxed{ \bf{ \: \dfrac{d}{dx} k = 0}}$

$\boxed{ \bf{ \: \dfrac{d}{dx} sinx = cosx}}$

$\boxed{ \bf{ \: \dfrac{d}{dx} cosx = - sinx}}$

$\boxed{ \bf{ \: \dfrac{d}{dx} tanx = {sec}^{2} x}}$

$\boxed{ \bf{ \: \dfrac{d}{dx} secx = secxtanx}}$

$\boxed{ \bf{ \: \dfrac{d}{dx} {a}^{x} = {a}^{x} log(a)}}$