CBSE BOARD X, asked by gouravgupta65, 6 months ago

असम के पर्यटन मंत्रालय द्वारा पर्यटक को आकर्षित करने के लिए असम की एतिहासिक विरासत का एक आकर्षक विज्ञापन तैयार कीजिए जिससे वहाँ के पर्यटन को बढ़ावा मिले |


please give right answer ​

Answers

Answered by abhimanyusikarwar200
2

Answer:

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Answered by TrustedAnswerer19
26

Topic :- This is your punishment for spamming in my question.

Differentiation

Given :-

\sf{y=\dfrac{x+1}{(x+2)^2}}

To Find :-

\sf {\dfrac{dy}{dx}\:for\:given\:'y'.}

Solution :-

\sf{y=\dfrac{x+1}{(x+2)^2}}

\sf {\dfrac{dy}{dx}=\dfrac{d}{dx}\left[\dfrac{x+1}{(x+2)^2}\right]}

\sf{\dfrac{(x+2)^2\cdot\dfrac{d(x+1)}{dx}-(x+1)\cdot\dfrac{d(x+2)^2}{dx}}{\{(x+2)^2\}^2}}

\sf {\left( \because \dfrac{d}{dx}\left(\dfrac{f}{g}\right)=\dfrac{g\cdot\dfrac{df}{dx}-f\cdot\dfrac{dg}{dx}}{g^2}\right)}

\sf{\dfrac{(x+2)^2\cdot\left(\dfrac{dx}{dx}+\dfrac{d(1)}{dx}\right)-(x+1)\cdot\dfrac{d(x+2)^2}{dx}}{(x+2)^4}}

\sf {\left(\because \dfrac{d(f\pm g)}{dx}=\dfrac{df}{dx}\pm\dfrac{dg}{dx}\right)}

\sf{\dfrac{(x+2)^2\cdot\left(1+0\right)-(x+1)\cdot\dfrac{d(x+2)^2}{dx}}{(x+2)^4}}

\sf {\left(\because \dfrac{dx}{dx}=1 \right)}

\sf {\left(\because \dfrac{dk}{dx}=0,when\:k\:is\:constant. \right)}

\sf{\dfrac{(x+2)^2\cdot1-(x+1)\cdot2(x+2)^{2-1}\cdot\dfrac{d(x+2)}{dx}}{(x+2)^4}}

\sf {\left(\because \dfrac{d(x+k)^n}{dx}=n(x+k)^{n-1}\cdot \dfrac{d(x+k)}{dx} \right)}

\sf{\dfrac{(x+2)^2-2(x+1)(x+2)\cdot\left(\dfrac{dx}{dx}+\dfrac{d(2)}{dx}\right)}{(x+2)^4}}

\sf {\left(\because \dfrac{d(f\pm g)}{dx}=\dfrac{df}{dx}\pm\dfrac{dg}{dx}\right)}

\sf{\dfrac{(x+2)^2-2(x+1)(x+2)\cdot\left(1+0\right)}{(x+2)^4}}

\sf {\left(\because \dfrac{dx}{dx}=1 \right)}

\sf {\left(\because \dfrac{dk}{dx}=0,when\:k\:is\:constant. \right)}

\sf{\dfrac{(x+2)^2-2(x+1)(x+2)}{(x+2)^4}}

\sf{\dfrac{x^2+4x+4-2(x^2+3x+2)}{(x+2)^4}}

\sf{\dfrac{x^2+4x+4-2x^2-6x-4}{(x+2)^4}}

\sf{\dfrac{x^2-2x^2+4x-6x+4-4}{(x+2)^4}}

\sf{\dfrac{-x^2-2x}{(x+2)^4}}

\sf{\dfrac{-x(x+2)}{(x+2)^4}}

\sf{\dfrac{-x}{(x+2)^3}}

Answer :-

\underline{\boxed{\sf{\dfrac{d}{dx}\left[\dfrac{x+1}{(x+2)^2}\right]=\dfrac{-x}{(x+2)^3}}}}

Topic :-

Indefinite Integration

To Solve :-

\sf{I=\displaystyle \int e^{2x}\sin 3x\:dx}

Concept Used :-

Integration by Parts

\sf{\displaystyle \int uv \:dx = u\int v\:dx - \int \left(\dfrac{du}{dx}\int v\:dx \right)dx}

where u and v are differentiable functions.

u and v are generally chosen as per the order of the letters in ILATE, where

I represents Inverse Trigonometric Function

L represents Logarithmic Function

A represents Algebraic Function

T represents Trigonometric Function and

E represents Exponential Function

Solution :-

\sf{I=\displaystyle \int e^{2x}\sin 3x\:dx}

\sf{Taking\:u=\sin3x \:and\:v=e^{2x}\:as\:per\:ILATE.}

Applying Integration by parts,

\sf{I=\displaystyle\sin3x\int e^{2x}\:dx - \int \left(\dfrac{d(\sin 3x)}{dx}\int e^{2x}\:dx \right)dx}

\sf{I=\displaystyle\sin3x\cdot\dfrac{e^{2x}}{2} - \int \left(\dfrac{d(\sin 3x)}{dx}\cdot\dfrac{e^{2x}}{2}\right)dx}

\sf{\left(\because\displaystyle \int e^{ax}\:dx=\dfrac{e^{ax}}{a}+C \right)}

\sf{\displaystyle I=\sin3x\cdot\dfrac{e^{2x}}{2} - \int \left(\cos3x\cdot\dfrac{d(3x)}{dx}\cdot\dfrac{e^{2x}}{2}\right)dx}

\sf{\left(\because \dfrac{d(\sin t)}{dx}=\cos t\cdot\dfrac{dt}{dx} \right)}

\sf{I = \displaystyle\sin3x\cdot\dfrac{e^{2x}}{2} - \int \left(3\cos3x\cdot\dfrac{e^{2x}}{2}\right)dx}

\sf{\left(\because \dfrac{d(kx)}{dx}=k,when\:k\:is\:a\:constant. \right)}

\sf{I=\displaystyle\sin3x\cdot\dfrac{e^{2x}}{2} - \dfrac{3}{2}\int \cos3x\cdot e^{2x}\:dx}

\sf{\left(\because \displaystyle \int k\cdot f(x)\:dx=k\int f(x)\:dx,when\:k\:is\:a\:constant. \right)}

\sf{Take\:\displaystyle\int \cos3x\cdot e^{2x}\:dx=J}

\sf{I=\displaystyle \sin3x\cdot\dfrac{e^{2x}}{2} - \dfrac{3}{2}J}

Solving J,

\sf{J=\displaystyle \int \cos3x\cdot e^{2x}\:dx}

\sf{Taking\:u=\cos3x \:and\:v=e^{2x}\:as\:per\:ILATE.}

Applying Integration by parts,

\sf {J=\displaystyle\cos3x\int e^{2x}\:dx - \int \left(\dfrac{d(\cos 3x)}{dx}\int e^{2x}\:dx \right)dx}

\sf{J=\displaystyle \cos3x\cdot\dfrac{e^{2x}}{2} - \int \left(\dfrac{d(\cos 3x)}{dx}\cdot\dfrac{e^{2x}}{2}\right)dx}

\sf{J=\displaystyle \cos3x\cdot\dfrac{e^{2x}}{2} - \int \left(-\sin3x\cdot\dfrac{d(3x)}{dx}\cdot\dfrac{e^{2x}}{2}\right)dx}

\sf{\left(\because \dfrac{d(\cos t)}{dx}=-\sin t\cdot\dfrac{dt}{dx} \right)}

\sf{J=\displaystyle \cos3x\cdot\dfrac{e^{2x}}{2} +\int \left(3\sin3x\cdot\dfrac{e^{2x}}{2}\right)dx}

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