Question

ate:
Time:
bject : :
পৰীয় ন মনােবিজ্ঞানের জনক কে ?​

Answer 1

Answer:

please question to pura likho

Answer 2

Answer:

Explanation:

EXPLANATION.

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

sin(x)

+(sin(x))

tan(x)

\begin{gathered}\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) } \end{gathered}

⟹lety=u+v

⟹

dx

dy

=

dx

du

+

dx

dv

⟹u=(tan(x))

sin(x)

⟹v=(sin(x))

tan(x)

\begin{gathered}\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) ) \end{gathered}

⟹u=(tan(x))

sin(x)

⟹takinglogonbothsidesweget

⟹ln(u)=ln(tan(x))

sin(x)

⟹

u

1

kya dekh raha hain chutiyen

dx

du

=sin(x)ln(tan(x))

\begin{gathered}\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 \impliemadarchods \: \frac{du}{dx} \: = (\tan(x) ) {}^{ \sin(x) }\Bigg[ \sec(x) \: + ln( \tan(x) ) \cos(x) \Bigg]\end{gathered}

⟹

u

1

dx

du

=sin(x).

tan(x)

sec

2

(x)

+ln(tan(x))cos(x)

⟹

dx

du

=u[sec(x)+ln(tan(x))cos(x)]

⟹

dx

du

=(tan(x))

sin(x)

[sec(x)+ln(tan(x))cos(x)]

\begin{gathered} \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) ) \end{gathered}

⟹v=(sin(x))

tan(x)

⟹takinglogonbothsidesweget

⟹

v

1

.

dx

dv

=ln(sin(x))

tan(x)

⟹

v

1

.

dx

dv

=tan(x)ln(sin(x))

\begin{gathered} \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] \end{gathered}

⟹

v

1

.

dx

dv

=tan(x).

sin(x)

cos(x)

+ln(sin(x))sec

2

(x)

⟹

dx

dv

=v[1+ln(sin(x))sec

2

(x)]

⟹

dx

dv

=sin(x)

tan(x)

[1+ln(sin(x))sec

2

(x)]

\begin{gathered} \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] \end{gathered}

⟹y=u+v

⟹addthevalueofuandv

⟹tan(x)

sin(x)

[sec(x)+ln(tan(x))cos(x)]+sin(x)

tan(x)

[1+ln(sin(x))sec

2

(x)]

Explanation:

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