Given that y = (cosx^ )^(sinx) find (dy/dx) and lim


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Question Number 64872 by Rio Michael last updated on 22/Jul/19

$G i v e n t h a t$ $y = {({c o s x}^{})}^{s i n x} f i n d \frac{d y}{d x}$ $a n d$ $\underset{x \to 0}{l i m} y$
Commented by kaivan.ahmadi last updated on 22/Jul/19

$l n y = s i n x l n (c o s x) \Rightarrow \frac{y^{'}}{y} = c o s x l n (c o s x) - \frac{s i n x}{c o s x} s i n x \Rightarrow$ $\frac{d y}{d x} = {(c o s x)}^{s i n x} (c o s x l n (c o s x) - \frac{{s i n}^{2} x}{c o s x})$ ${l i m}_{x \to 0} {c o s x}^{s i n x} = 1^{0} = 1$
Commented by mathmax by abdo last updated on 22/Jul/19
$we have y =e^(sinxln(cosx)) ⇒(dy/dx) =(d/dx)( sinx ln(cosx))e^(sinxln(cosx)) ={cosxln(cosx) +sinx ((−sinx)/(cosx))}y(x) ⇒ y^′ (x)=((cos^2 x ln(cosx)−sin^2 x)/(cosx))× (cosx)^(sinx)$
$w e h a v e y = e^{s i n x l n (c o s x)} \Rightarrow \frac{d y}{d x} = \frac{d}{d x} (s i n x l n (c o s x)) e^{s i n x l n (c o s x)}$ $= {c o s x l n (c o s x) + s i n x \frac{- s i n x}{c o s x}} y (x) \Rightarrow$ $y^{^{'}} (x) = \frac{{c o s}^{2} x l n (c o s x) - {s i n}^{2} x}{c o s x} \times {(c o s x)}^{s i n x}$
Commented by mathmax by abdo last updated on 22/Jul/19

$w e h a v e y (x) = e^{s i n x l n (c o s x)} a n d {l i m}_{x \to 0} s i n x l n (c o s x) = 0 \Rightarrow$ ${l i m}_{x \to 0} y (x) = 1$
Commented by Masumsiddiqui399@gmail.com last updated on 23/Jul/19
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