বিপরীত ফাংশন ও পরামিতিক ফাংশনের অন্তরজ
যদি f(x)= sinx হয় তবে f(cos-1x) এর অন্তরজ কোনটি?
1(21−x2 \frac{1}{\left ( 2 \sqrt{1 - x^{2}} \right.} (21−x21
−1(21−x2 - \frac{1}{\left ( 2 \sqrt{1 - x^{2}} \right.} −(21−x21
x(1−x2 \frac{x}{\left ( \sqrt{1 - x^{2}} \right.} (1−x2x
−x(1−x2 - \frac{x}{\left ( \sqrt{1 - x^{2}} \right.} −(1−x2x
f(x)=sinx ^{\mathrm{}} \mathrm{f}(\mathrm{x})=\sin \mathrm{x} f(x)=sinx
f(cos−1x)=sin(cos−1x) f\left(\cos ^{-1} x\right)=\sin \left(\cos ^{-1} x\right) f(cos−1x)=sin(cos−1x)
=sin(sin−11−x2)=1−x2∴f′(cos−1x)=−2x21−x2 \begin{array}{r} \quad=\sin \left(\sin ^{-1} \sqrt{1-x^{2}}\right)=\sqrt{1-x^{2}} \\ \therefore \quad f^{\prime}\left(\cos ^{-1} x\right)=\frac{-2 x}{2 \sqrt{1-x^{2}}} \text { } \end{array} =sin(sin−11−x2)=1−x2∴f′(cos−1x)=21−x2−2x
If u=f(x2),v=g(x3),f(x)=sinx,g1(x)=cosxu=f(x^{2}), v=g(x^{3}),f(x)=\sin x, g^{1}(x)=\cos xu=f(x2),v=g(x3),f(x)=sinx,g1(x)=cosx then find dudv\frac{du}{dv}dvdu
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?
x= at2 এবং y=2at
dydx∣t=1 \frac{dy}{dx} |_{t = 1} dxdy∣t=1 এর মান কত ?