differenciatethefollowing i)x^x +(sinx)^(lnx) = ii)sin^(−1) (tanhx)= iii)(√(1+x^2 /1−x^2 =))


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Question Number 54790 by Anju last updated on 11/Feb/19

$d i f f e r e n c i a t e t h e f o l l o w i n g$ $i) x^{x} + {(s i n x)}^{l n x} =$ $i i) {s i n}^{- 1} (t a n h x) =$ $i i i) \sqrt{1 + x^{2} / 1 - x^{2} =}$
Commented by maxmathsup by imad last updated on 11/Feb/19

$1) l e t f (x) = x^{x} + {(s i n x)}^{l n (x)} \Rightarrow f (x) = e^{x l n (x)} + e^{l n (x) l n (s i n x)} w i t h x > 0 \Rightarrow$ $f^{^{'}} (x) = {(x l n x)}^{^{'}} x^{x} + {(l n x l n (s i n x))}^{^{'}} {(s i n x)}^{l n (x)}$ $= (l n x + 1) x^{x} + (\frac{l n (s i n x)}{x} + l n x \frac{c o s x}{s i n x}) {(s i n x)}^{l n (x)}$ $= (1 + l n (x)) x^{x} + (\frac{l n (s i n x)}{x} + l n x c o t a n (x)) {(s i n x)}^{l n (x)} .$
Commented by maxmathsup by imad last updated on 11/Feb/19

$i i) l e t g (x) = a r c s i n (t a n h x) w e h a v e t a n h x = \frac{s h (x)}{c h (x)} = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} = \frac{e^{2 x} - 1}{e^{2 x} + 1} \Rightarrow$ $g (x) = a r c s i n (\frac{e^{2 x} - 1}{e^{2 x} + 1}) \Rightarrow g^{^{'}} (x) = \frac{{(\frac{e^{2 x} - 1}{e^{2 x} + 1})}^{^{'}}}{\sqrt{1 - {(\frac{e^{2 x} - 1}{e^{2 x} + 1})}^{2}}} b u t$ ${(\frac{e^{2 x} - 1}{e^{2 x} + 1})}^{^{'}} = \frac{2 e^{2 x} (e^{2 x} + 1) - 2 e^{2 x} (e^{2 x} - 1)}{{(e^{2 x} + 1)}^{2}} = \frac{4 e^{2 x}}{{(e^{2 x} + 1)}^{2}} \Rightarrow$ $g^{^{'}} (x) = \frac{4 e^{2 x}}{{(e^{2 x} + 1)}^{2} \frac{\sqrt{{(e^{2 x} + 1)}^{2} - {(e^{2 x} - 1)}^{2}}}{e^{2 x} + 1}} = \frac{4 e^{2 x}}{(e^{2 x} + 1) \sqrt{e^{4 x} + 2 e^{2 x} + 1 - e^{4 x} + 2 e^{2 x} - 1}}$ $= \frac{4 e^{2 x}}{(1 + e^{2 x}) 2 e^{x}} = \frac{2 e^{x}}{(1 + e^{2 x})} \Rightarrow g^{^{'}} (x) = \frac{2 e^{x}}{(1 + e^{2 x})} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 11/Feb/19

	$1) y = x^{x} + {(s i n x)}^{l n x}$ $y = u + v$ $\frac{d y}{d x} = \frac{d u}{d x} + \frac{d v}{d x}$ $u = x^{x}$ $l n u = x l n x$ $\frac{1}{u} \frac{d u}{d x} = x \times \frac{1}{x} + l n x \times 1$ $\frac{d u}{d x} = x^{x} (1 + l n x)$ $v = {(s i n x)}^{l n x}$ $l n v = l n x l n (s i n x)$ $\frac{1}{v} \times \frac{d v}{d x} = l n x \times \frac{1}{s i n x} \times c o s x + l n (s i n x) \times \frac{1}{x}$ $\frac{d v}{d x} = {(s i n x)}^{l n x} [c o t x l n x + \frac{l n (s i n x)}{x}]$ $s o a n s w e r i s$ $\frac{d u}{d x} + \frac{d v}{d x}$ $= x^{x} (1 + l n x) + {(s i n x)}^{l n x} [c o t x l n x + \frac{l n (s i n x)}{x}]$
	Answered by tanmay.chaudhury50@gmail.com last updated on 11/Feb/19
	$iii)y=(√((1+x^2 )/(1−x^2 ))) lny=(1/2){ln(1+x^2 )−ln(1−x^2 )} (1/y)(dy/dx)=(1/2){((2x)/(1+x^2 ))−((−2x)/(1−x^2 ))} (1/y)(dy/dx)=x{((1−x^2 +1+x^2 )/(1−x^4 ))} (dy/dx)=(√((1+x^2 )/(1−x^2 ))) (((2x)/(1−x^4 )))$
	$i i i) y = \sqrt{\frac{1 + x^{2}}{1 - x^{2}}}$ $l n y = \frac{1}{2} {l n (1 + x^{2}) - l n (1 - x^{2})}$ $\frac{1}{y} \frac{d y}{d x} = \frac{1}{2} {\frac{2 x}{1 + x^{2}} - \frac{- 2 x}{1 - x^{2}}}$ $\frac{1}{y} \frac{d y}{d x} = x {\frac{1 - x^{2} + 1 + x^{2}}{1 - x^{4}}}$ $\frac{d y}{d x} = \sqrt{\frac{1 + x^{2}}{1 - x^{2}}} (\frac{2 x}{1 - x^{4}})$
	Answered by tanmay.chaudhury50@gmail.com last updated on 11/Feb/19

	$i i) y = {s i n}^{- 1} (t a n h x)$ $\frac{d y}{d x} = \frac{1}{\sqrt{1 - {(t a n h x)}^{2}}} \times {s e c h}^{2} x$
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