Question

consider the following equation :-
$\\ \bullet \quad \sf {x}^{y} = {e}^{x - y}$

– Find dy/dx at x = 1

– Find d²y/dx² at x = 1
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Answer 1

Given-

$x^y = {e}^{x - y}$

Taking natural log on both sides-

$\sf log_{e} {x}^{y} = log_{e} {e}^{ (x - y) }$

$\sf \: y \: log \: x = (x-y)log_e e$

$\quad\quad\quad\quad\quad \: \quad ( log_e e = 1)$

$\sf y \: log x = x-y$

$\sf y \: log x +y = x$

$\sf y(log x + 1) = x$

$\sf y = \large{\frac{x} { log \: x + 1}}$

Differentiating Both Sides w.r.t x -

$\sf \: \frac{dy}{dx} = \large\frac{(1+logx )\frac{dx}{dx}- x.(\frac{d(1+logx)}{dx})}{(log x +1)^2}$

$\bf \quad\quad\quad\quad\quad(by \: division \: rule \: in \: differentiation)$

$\sf \frac{dy}{dx} = \large\frac{(1+log \: x ).1 - x.(0+ \frac{1}{x})}{(log \: x +1)^2}$

$\sf \frac{dy}{dx} = \large\frac{(1+log \: x ) - x.( \frac{1}{x})}{(log \: x +1)^2}$

$\sf \frac{dy}{dx} = \large\frac{\cancel1+logx - \cancel{1}}{(log x +1)^2}$

$\boxed{ \sf \frac{dy}{dx} = \large\frac{log \: x }{(log \: x + 1)^2}}$

$\underline{\underline {\bf\purple{dy/dx \: at \:x = 1\:}}}$

$\rm (\frac{dy}{dx})_{x=1} =\frac{log \: 1}{(log \: 1+ 1)^2}$

$\rm (\frac{dy}{dx})_{x=1} =\frac{0}{(0+ 1)^2}$

$\rm (\frac{dy}{dx})_{x=1} =\frac{0}{( 1)^2}$

$\red{\boxed{\rm(\frac{dy}{dx})_{x=1} =0}}$

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Now,

$\rm \frac{dy}{dx}=\large\frac{log \: x}{(log \: x+ 1)^2}$

Differentiating both sides w.r.t x -

$\rm \frac{d}{dx}(\frac{dy}{dx}) = \frac{d}{dx}\: [\frac{log\: x}{(log\:x+1)^2}]$

$\rm \frac{d^2y}{dx^2} = \frac{(log\: x+1)^2.\frac{d}{dx}(log \:x)-log\:x.\frac{d}{dx}(log\:x +1)^2}{[(log\:x+1)^2]^2}$

$\bf \quad\quad\quad\quad\quad(by \: division \: rule \: in \: differentiation)$

$\rm \frac{d^2y}{dx^2} = \frac{(log\: x+1)^2.\frac{1}{x}-log\:x. \: 2 \times (log\:x +1)^{2-1}.\frac{d}{dx}(log\:x +1)}{[(log\:x+1)^2]^2}$

$\rm \frac{d^2y}{dx^2} = \frac{(log\: x+1)^2.\frac{1}{x}-2.log\:x.(log\:x +1).(\frac{1}{x}+0)}{[(log\:x+1)^2]^2}$

$\rm \frac{d^2y}{dx^2} = \frac{(log\: x+1)^2.\frac{1}{x}-2.log\:x.(log\:x +1).\frac{1}{x}}{[(log\:x+1)^2]^2}$

$\rm \frac{d^2y}{dx^2} = \frac{(\frac{log\: x +1}{x})(log\:x +1-2.log\:x)}{[(log\:x+1)^2]^2}$

$\underline{\underline {\bf\red{d^2y/dx^2 \: at \:x = 1\:}}}$

$\rm (\frac{d^2y}{dx^2})_{x=1} = \frac{(\frac{log\: 1+1}{1})(log\:1 +1-2.log\:1)}{[(log\:1+1)^2]^2}$

$\rm (\frac{d^2y}{dx^2})_{x=1} = \frac{(\frac{0+1}{1})(0 +1-2\times 0)}{[(0+1)^2]^2}$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \bf(log\: 1 =0)$

$\rm (\frac{d^2y}{dx^2})_{x=1} = \frac{(\frac{1}{1}).(1-0)}{(1)^4}$

$\rm (\frac{d^2y}{dx^2})_{x=1} = \frac{1}{1}$

$\pink{\boxed{\rm(\frac{d^2y}{dx^2})_{x=1} = 1}}$

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Remember-

$\tt log \: {x}^{n} = n \: log \: x$

Division rule of Differentiation -

$\tt \large\frac{d}{dx} [\frac{f(x)}{g(x)} ] = \frac{g(x).f'(x) - f(x).g'(x) }{(g(x))^2}$

$\implies$f'(x) and g'(x) are differentiated terms

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Answer 2

Refer the attachment for your answer (◕દ◕)

Hope it helps & my writing is easily Understandable ＼(^o^)／

If needed then I will write the answer in latex if writing is not understandable but I tried my best to write a writing that it will be easy to understand for you ( ꈍᴗꈍ )