বিপরীত ফাংশন ও পরামিতিক ফাংশনের অন্তরজ
lf the parametric equation of a curve given by x=etcost, y=etsintx=e^{t}\cos t,\ y=e^{t}\sin tx=etcost, y=etsint, then the tangent to the curve at the point t=π4t=\dfrac{\pi}{4}t=4π makes with axis of xxx the angle.
000
π4\dfrac{\pi}{4}4π
π3\dfrac{\pi}{3}3π
π2\dfrac{\pi}{2}2π
at t=π/4at\ t={\pi}/{4}at t=π/4
y=etsinty=e^{t}sinty=etsint
dydl=et(sin t+cos t)\dfrac{dy}{dl}=e^{t}(sin\ t+cos\ t)dldy=et(sin t+cos t)
dxdl=dt(cos t+sin t)\dfrac{dx}{dl}=d^{t}(cos\ t+sin\ t)dldx=dt(cos t+sin t)
dydx=sin t+cos tcos t−sin t\dfrac{dy}{dx}=\dfrac{sin \ t+cos\ t}{cos\ t-sin\ t}dxdy=cos t−sin tsin t+cos t
=α= \alpha=α
∴tan−1(∞)=π/2\therefore tan^{-1}(\infty)={\pi}/{2}∴tan−1(∞)=π/2
Let the function y=f(x)y=f(x)y=f(x) be given by x=t5−5t3−20t+7x=t^{5}-5t^{3}-20t+7x=t5−5t3−20t+7 and y=4t3−3t2−18t+3y=4t^{3}-3t^{2}-18t+3y=4t3−3t2−18t+3, where tϵ(−2,2)t\epsilon \left ( -2, 2 \right )tϵ(−2,2). Then f′(x)f^{'}(x)f′(x) at t=1t=1t=1 is ?
x=a(θ-sinθ), y=a(1-cosθ) হলে -
নিচের কোনটি সঠিক?
y=sin−1[4x1+4x] y = \sin^{- 1}{\left [ \frac{4 \sqrt{x}}{1 + 4 x} \right ]} y=sin−1[1+4x4x] হলে, (dydx)((4,2) \left ( \frac{dy}{dx} \right )_{\left ( \left ( 4 , 2 \right ) \right.} (dxdy)((4,2) এর মান কত?
If for x∈(0,14)x \in \left(0, \dfrac{1}{4}\right)x∈(0,41), the derivative tan−1(6xx1−9x3)\tan^{-1} \left(\dfrac{6x\sqrt{x}}{1-9x^{3}}\right)tan−1(1−9x36xx) is x.g(x)\sqrt{x}.g(x)x.g(x), then g(x)g(x)g(x) equals :