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উচ্চতর গণিত প্রথম পত্র( Class11-12)

$\displaystyle \frac{dy}{dx}$=$\displaystyle \frac{d}{dx}$$ $$\displaystyle \left(\frac{1+sinx}{1-sinx}\right)$ =$\displaystyle \frac{( 1-sinx)\frac{d}{dx}( 1+sinx) -( 1+sinx)\frac{d}{dx}( 1-sinx)}{( 1-sinx)^{2}}$ =$\displaystyle \frac{( 1-sinx)\frac{d}{dx}( 1) +\frac{d}{dx}( sinx) -( 1+sinx)\frac{d}{dx}( 1) -\frac{d}{dx}( sinx)}{( 1-sinx)^{2}}$ =$\displaystyle \frac{( 1-sinx)(...

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)$\displaystyle \frac{sinx}{x+cosx}$ মনে করি, y=$\displaystyle \frac{sinx}{x+cosx}$ x এর সাপেক্ষে অন্তরীকরণ করে পাই, =$\displaystyle \frac{( x+cosx)\frac{d}{dx}( sinx) -sinx\frac{d}{dx}( x+cosx)}{( x+cosx)^{2}}$ =$\displaystyle \frac{( x+cosx) .cosx-sinx\frac{d}{dx}( x) +\frac{d}{dx}( cosx)}{( x+cosx)^{2}}$ =$\displaystyle...

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i) $\displaystyle \frac{1-cotx}{1+cotx}$ মনে করি, y=$\displaystyle \frac{1-cotx}{1+cotx}$ x এর সাপেক্ষে অন্তরীকরণ করে পাই, $\displaystyle \frac{dy}{dx}$=$\displaystyle \frac{d}{dx}$$\displaystyle \left(\frac{1-cotx}{1+cotx}\right)$ =$\displaystyle \frac{( 1+cotx)\frac{d}{dx}( 1-cotx) -( 1-cotx)\frac{d}{dx}( 1+cotx)}{( 1+cotx)^{2}}$ = $\displaystyle \frac{(...

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i) $\displaystyle \frac{lnx}{cosx}$ মনে করি, y=$\displaystyle \frac{lnx}{cosx}$ =$\displaystyle \frac{1}{cosx}$$\displaystyle lnx$ =lnx secx x এর সাপেক্ষে অন্তরীকরণ করে পাই, $\displaystyle \frac{dy}{dx}$=$\displaystyle \frac{d}{dx}$(lnx secx) =lnx$\displaystyle \frac{d}{dx}$(secx)+(secx)$\displaystyle \frac{d}{dx}$(lnx) =lnx.secx.tanx+(secx).$\displaystyle \frac{1}{x}$ =$\displaystyle...

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ii) $\displaystyle \frac{sinx+cosx}{\sqrt{1+sin2x}}$ মনে করি, y=$\displaystyle \frac{sinx+cosx}{\sqrt{1+sin2x}}$ =$\displaystyle \frac{sinx+cosx}{\sqrt{sin^{2} x+cos^{2} x+2sinx\ cosx}}$ = $\displaystyle \frac{( sinx+cosx)}{\sqrt{( sinx+cosx)^{2}}}$ = $\displaystyle \frac{( sinx+cosx)}{( sinx+cosx)}$ =1 x এর সাপেক্ষে অন্তরীকরণ করে...

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ii) x$\displaystyle ^{lnx}$ মনে করি, y=x$\displaystyle ^{lnx}$ (উভয়পক্ষ ln নিয়ে পাই) lny=lnx$\displaystyle ^{lnx}$ lny=lnx.lnx x এর সাপেক্ষে অন্তরীকরণ করে পাই, $\displaystyle \frac{dy}{dx}$=y$\displaystyle \left$ $\displaystyle \frac{dy}{dx}$=x$\displaystyle ^{lnx}$($\displaystyle \frac{2lnx}{x}$) (ans):...

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ii) x$\displaystyle ^{\frac{1}{x}}$ মনে করি, y=x$\displaystyle ^{\frac{1}{x}}$ lny=$\displaystyle \frac{1}{x}$ lnx (উভয়পক্ষ ln নিয়ে পাই) x এর সাপেক্ষে অন্তরীকরণ করে পাই, =x$\displaystyle ^{\frac{1}{x}}$.$\displaystyle \frac{1}{x^{2}}$(1-lnx) =x$\displaystyle ^{\frac{1}{x}}$$\displaystyle ^{-2}$(1-lnx) (ans):

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ii) $\displaystyle \left{\frac{sinx-xcosx}{xsinx+cosx}\right}$ মনে করি, y=$\displaystyle \left{\frac{sinx-xcosx}{xsinx+cosx}\right}$ x এর সাপেক্ষে অন্তরীকরণ করে পাই, $\displaystyle \frac{dy}{dx}$=$\displaystyle \frac{d}{dx}$$\displaystyle \left{\frac{sinx-xcosx}{xsinx+cosx}\right}$ =$\displaystyle \frac{( xsinx+cosx)\frac{d}{dx}( sinx-xcosx) -( sinx-xcosx)\frac{d}{dx}( xsinx+cosx)}{( xsinx+cosx)^{2}}$ =$\displaystyle \frac{( xsinx+cosx)\frac{d}{dx}(...

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iv) 2cosec 2xcos{ln(tanx)} মনে করি, y=2cosec 2xcos{ln(tanx)} x এর সাপেক্ষে অন্তরীকরণ করে পাই, $\displaystyle \frac{dy}{dx}$=$\displaystyle \frac{d}{dx}$ 2cosec 2xcos{ln(tanx)} =2 =2[cosec2x.(- sin{ln(tanx)}$\displaystyle \frac{d}{dx}$ln(tanx)+cos{ln(tanx)}(-cosec2x.cot2x$\displaystyle \frac{d}{dx}$(2x) =2[cosec2x.(- sin{ln(tanx)}.$\displaystyle \frac{1}{tanx}$.$\displaystyle...

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iv) cos(lnx)+ln(tanx) মনে করি, y=cos(lnx)+ln(tanx) $\displaystyle \frac{dy}{dx}$=$\displaystyle \frac{d}{dx}$cos(lnx)+ln(tanx) =$\displaystyle \frac{d}{dx}$cos(lnx)+$\displaystyle \frac{d}{dx}$ln(tanx) = - sin(lnx)$\displaystyle \frac{d}{dx}$(lnx)+$\displaystyle \frac{1}{tanx}$$\displaystyle \frac{d}{dx}$(tanx) =- sin(lnx).$\displaystyle \frac{1}{x}$+$\displaystyle \frac{1}{tanx}$.sec$\displaystyle ^{2}$ = $\displaystyle -\ \frac{sin( lnx)}{x}$+$\displaystyle \frac{sec^{2}}{tanx}$...

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