ত্রিকোনমিতিক ফাংশনের অন্তরজ
xxx এর সাপেক্ষে অন্তরক সহগ নিচের কোনটি? sin−1(tan−1x) \sin ^{-1}\left(\tan ^{-1} x\right) sin−1(tan−1x)
−1(1+x2)1−(tan−1x)2 -\frac{1}{\left(1+x^{2}\right) \sqrt{1-\left(\tan ^{-1} x\right)^{2}}}−(1+x2)1−(tan−1x)21
1(1+x2)1−(tan−1x)2 \frac{1}{\left(1+x^{2}\right) \sqrt{1-\left(\tan ^{-1} x\right)^{2}}}(1+x2)1−(tan−1x)21
1(1−x2)1−(tan−1x)2 \frac{1}{\left(1-x^{2}\right) \sqrt{1-\left(\tan ^{-1} x\right)^{2}}}(1−x2)1−(tan−1x)21
1(1+x2)1+(tan−1x)2 \frac{1}{\left(1+x^{2}\right) \sqrt{1+\left(\tan ^{-1} x\right)^{2}}}(1+x2)1+(tan−1x)21
Solve:
ddx{sin−1(tan−1x)}=11−(tan−1x)2ddx(tan−1x) \begin{array}{l} \frac{d}{d x}\left\{\sin ^{-1}\left(\tan ^{-1} x\right)\right\} \\ =\frac{1}{\sqrt{1-\left(\tan ^{-1} x\right)^{2}}} \frac{d}{d x}\left(\tan ^{-1} x\right) \end{array} dxd{sin−1(tan−1x)}=1−(tan−1x)21dxd(tan−1x)
=11−(tan−1x)211+x2=1(1+x2)1−(tan−1x)2 \begin{array}{l} =\frac{1}{\sqrt{1-\left(\tan ^{-1} x\right)^{2}}} \frac{1}{1+x^{2}} \\ =\frac{1}{\left(1+x^{2}\right) \sqrt{1-\left(\tan ^{-1} x\right)^{2}}} \end{array} =1−(tan−1x)211+x21=(1+x2)1−(tan−1x)21
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-
নিচের কোনটি সঠিক?
