1)find (dy/dx) y=(sin(x))^(cos^(−1) (x)) 2)lim


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Question Number 97108 by M±th+et+s last updated on 06/Jun/20

$1) f i n d \frac{d y}{d x}$ $y = {(s i n (x))}^{{c o s}^{- 1} (x)}$ $2) {\underset{x \to \infty}{l i m x}}^{\frac{1}{x}}$
Commented by M±th+et+s last updated on 06/Jun/20

$t h a n k s f o r s o l u t i o n s$
	Answered by Sourav mridha last updated on 06/Jun/20

	$1) \frac{d y}{d x} = \frac{d}{d x} e^{\cos^{- 1} (x) . l n s i n (x)}$ $= {(s i n (x))}^{\cos^{- 1} (x)} [c o t x . \cos^{- 1} (x) - \frac{l n (s i n (x))}{\sqrt{1 - x^{2}}}]$
	Answered by Sourav mridha last updated on 06/Jun/20
	$2)u=x^(1/x) ⇒ ⇒ln(u)=((ln(x))/x) ⇒lim_(x→∞) [ln(u)]=lim_(x→∞) [((ln(x))/x)]{(∞/∞) form using LHopital} =0 ⇒lim_(x→∞) [(u)]=e^0 =1$
	$2) u = x^{\frac{1}{x}} \Rightarrow$ $\Rightarrow l n (u) = \frac{l n (x)}{x}$ $\Rightarrow \lim_{x \to \infty} [l n (u)] = \lim_{x \to \infty} [\frac{l n (x)}{x}] {\frac{\infty}{\infty} f o r m u s i n g L H o p i t a l}$ $= 0$ $\Rightarrow \lim_{x \to \infty} [(u)] = e^{0} = 1$
	Answered by abdomathmax last updated on 06/Jun/20

	$1) y = {(sinx)}^{arcosx} \Rightarrow y = e^{arcosxln (sinx)}$ $\Rightarrow \frac{dy}{dx} = {(arcosxln (\sin))}^{^{'}} e^{arcosxln (sinx)}$ $= (- \frac{\ln (sinx)}{\sqrt{1 - x^{2}}} + arcosx \times cotanx) y (x)$
	Answered by abdomathmax last updated on 06/Jun/20

	$2) let f (x) = x^{\frac{1}{x}} \Rightarrow f (x) = e^{\frac{1}{x} lnx} \Rightarrow$ $\lim_{x \to + \infty} f (x) = \lim_{x \to + \infty} e^{\frac{lnx}{x}} = e^{o} = 1$
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