Question

Plzzz do its urgent........​

Answer 1

⭐Explanation ⭐

$\sf \implies \: y \: = ( \tan(x) ) {}^{ \sin(x) } \: + \: (\sin(x)) {}^{ \tan(x) }$

$\sf \implies \:let \: y \: = u \: + v \\ \\ \sf \implies \: \frac{dy}{dx} = \frac{du}{dx} \: + \: \frac{dv}{dx} \\ \\ \sf \implies \: \: u \: = (\tan(x)) {}^{ \sin(x) } \\ \\ \sf \implies \: \: v \: =( \sin(x)) {}^{ \tan(x) }$

$\sf \implies \:u \: = (\tan(x)) {}^{ \sin(x) } \\ \\ \sf \implies \: \: taking \: log \: on \: both \: sides \: we \: get \\ \\ \sf \implies \: ln(u) = ln( \tan(x) ) {}^{ \sin(x) } \\ \\ \sf \implies \: \frac{1}{u} \frac{du}{dx} = \sin(x) ln( \tan(x) )$

$\sf \implies \: \dfrac{1}{u} \dfrac{du}{dx} = \sin(x) . \dfrac{ \sec {}^{2} (x) }{ \tan(x) } \: + \: ln( \tan(x) ) \cos(x) \\ \\ \sf \implies \: \frac{du}{dx} = u \Bigg[ \sec(x) \: + \: ln( \tan(x) ) \cos(x) \Bigg] \\ \\ \sf \implies \: \frac{du}{dx} \: = (\tan(x) ) {}^{ \sin(x) }\Bigg[ \sec(x) \: + ln( \tan(x) ) \cos(x) \Bigg]$

$\sf \implies \: v \: =( \sin(x) ) {}^{ \tan(x) } \\ \\ \sf \implies \: taking \: log \: on \: both \: sides \: we \: get \\ \\ \sf \implies \: \frac{1}{v}. \frac{dv}{dx} = ln( \sin(x) ) {}^{ \tan(x) } \\ \\ \sf \implies \: \frac{1}{v} . \frac{dv}{dx} = \tan(x) ln( \sin(x) )$

$\sf \implies \: \dfrac{1}{v}. \dfrac{dv}{dx} = \tan(x). \dfrac{ \cos(x) }{ \sin(x) } \: + \: ln( \sin(x) ) \sec {}^{2} (x) \\ \\ \sf \implies \: \frac{dv}{dx} = v \Bigg[1 \: + \: ln( \sin(x) ) \sec {}^{2} (x) \Bigg] \\ \\ \sf \implies \: \frac{dv}{dx} \: = \sin(x) {}^{ \tan(x) }\Bigg[1 \: + \: ln( \sin(x) ) \sec {}^{2} (x) \Bigg]$

$\sf \implies \: y \: = u \: + v \: \\ \\ \sf \implies \: add \: the \: value \: of \: u \: and \: v \\ \\ \sf \implies \: \tan(x) {}^{ \sin(x) }\Bigg[ \sec(x) \: + \: ln( \tan(x) ) \cos(x) \Bigg] \: + \: \sin(x) {}^{ \tan(x) } \Bigg[ 1 \: + \: ln( \sin(x) ) \sec {}^{2} (x) \Bigg]$

Answer 2

Given:

• $\rm y = (\tan x)^{\sin x} + (\sin x)^t$

To find:

• $\rm \frac{dy}{dx}$

Solution:

$\therefore \rm y = (\tan x)^{\sin x} + (\sin x)^t$

$\implies \rm y = 1 + [(\tan x) + (\sin x)]^{\sin xt}$

$\implies \rm y = 1 + [(\tan x) + (\sin x)]^{\sin xt}$

$\because \bf \rm 1 + \tan A = \sin A$

$\implies \rm y = sin x + (\sin x)]^{\sin xt}$

$\implies \rm y = 1^{\sin xt}$

Therefore,

$\implies \frac{1^{\sin xt}}{x} = 1$

So,

Answer = 1.