ত্রিকোনমিতিক ফাংশনের অন্তরজ
যদি y=12(sin−1x)2 y=\frac{1}{2}\left(\sin ^{-1} x\right)^{2} y=21(sin−1x)2 হয়, তবে প্রমান কর যে (1−x2)y2−xy1=1 \left(1-\mathrm{x}^{2}\right) \mathrm{y}_{2}-\mathrm{xy}_{1}=1 (1−x2)y2−xy1=1
y1=12⋅2(sin−1x)ddx⋅(sin−1x)=sin−1x⋅11−x2 \mathrm{y}_{1}=\frac{1}{2} \cdot 2\left(\sin ^{-1} \mathrm{x}\right) \frac{\mathrm{d}}{\mathrm{dx}} \cdot\left(\sin ^{-1} \mathrm{x}\right)=\sin ^{-1} \mathrm{x} \cdot \frac{1}{\sqrt{1-\mathrm{x}^{2}}} y1=21⋅2(sin−1x)dxd⋅(sin−1x)=sin−1x⋅1−x21 বा, (1−x2)y1=sin−1x \left(\sqrt{1-\mathrm{x}^{2}}\right) \mathrm{y}_{1}=\sin ^{-1} \mathrm{x} (1−x2)y1=sin−1x
বা, (1−x2)y2+y1⋅(−2x)21−x2=11−x2∴(1−x2)y2−xy1=1 \left(\sqrt{1-\mathrm{x}^{2}}\right) \mathrm{y}_{2}+\mathrm{y}_{1} \cdot \frac{(-2 \mathrm{x})}{2 \sqrt{1-\mathrm{x}^{2}}}=\frac{1}{\sqrt{1-\mathrm{x}^{2}}} \therefore\left(1-\mathrm{x}^{2}\right) \mathrm{y}_{2}-\mathrm{xy}_{1}=1 \quad (1−x2)y2+y1⋅21−x2(−2x)=1−x21∴(1−x2)y2−xy1=1 (Proved)
If the functions f(x)=sin(x+a) \displaystyle f\left ( x \right )=\sin \left ( x+a \right ) f(x)=sin(x+a) and g(x)=bsinx+ccosx \displaystyle g\left ( x \right )=b\sin x+c\cos x g(x)=bsinx+ccosx satisfy f(0)=g(0) \displaystyle f\left ( 0 \right )=g\left ( 0 \right ) f(0)=g(0) and f′(0)=g′(0) \displaystyle {f}'\left ( 0 \right )={g}'\left ( 0 \right ) f′(0)=g′(0) then
dydx\displaystyle\frac{dy}{dx}dxdy at t=π4\displaystyle t=\frac{\pi}{4}t=4π for x=a[cost+12logtan2t2]\displaystyle x=a\left[\cos{t}+\frac{1}{2}\log{\tan^2{\frac{t}{2}}}\right]x=a[cost+21logtan22t] and y=asinty=a\sin{t}y=asint is
y=ln(cosx) y=\ln (\cos x) y=ln(cosx) হলে, dydx \frac{d y}{d x} dxdy এর মান কত?
y=sin(1x) y = \sin{\left ( \frac{1}{x} \right )} y=sin(x1) হলে dydx \frac{dy}{dx} dxdy এর মান-
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