বিপরীত ফাংশন ও পরামিতিক ফাংশনের অন্তরজ
xxx এর সাপেক্ষে অন্তরক সহগ নিচের কোনটি? sin−1x5 \sqrt{\sin ^{-1} x^{5}} sin−1x5
10x42sin−1x51−x10 \frac{10 x^{4}}{2 \sqrt{\sin ^{-1} x^{5}} \sqrt{1-x^{10}}}2sin−1x51−x1010x4
5x42sin−1x51−x5 \frac{5 x^{4}}{2 \sqrt{\sin ^{-1} x^{5}} \sqrt{1-x^{5}}}2sin−1x51−x55x4
5x42sinx51+x10 \frac{5 x^{4}}{2 \sqrt{\sin x^{5}} \sqrt{1+x^{10}}}2sinx51+x105x4
5x42sin−1x51−x10 \frac{5 x^{4}}{2 \sqrt{\sin ^{-1} x^{5}} \sqrt{1-x^{10}}}2sin−1x51−x105x4
Solve:ddx(⋅sin−1x5)=12sin−1x5ddx(sin−1x5) \frac{d}{d x}\left(\cdot \sqrt{\sin ^{-1} x^{5}}\right)=\frac{1}{2 \sqrt{\sin ^{-1} x^{5}}} \frac{d}{d x}\left(\sin ^{-1} x^{5}\right) dxd(⋅sin−1x5)=2sin−1x51dxd(sin−1x5)
=12sin−1x511−(x5)2ddx(x5)=12sin−1x51−x10(5x4)=5x42sin−1x51−x10 \begin{array}{l} =\frac{1}{2 \sqrt{\sin ^{-1} x^{5}}} \frac{1}{\sqrt{1-\left(x^{5}\right)^{2}}} \frac{d}{d x}\left(x^{5}\right) \\ =\frac{1}{2 \sqrt{\sin ^{-1} x^{5}} \sqrt{1-x^{10}}}\left(5 x^{4}\right) \\ =\frac{5 x^{4}}{2 \sqrt{\sin ^{-1} x^{5}} \sqrt{1-x^{10}}} \end{array} =2sin−1x511−(x5)21dxd(x5)=2sin−1x51−x101(5x4)=2sin−1x51−x105x4
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
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 ?
ddxtan−1(1−x1+x)=\displaystyle\frac{d}{dx}\tan^{-1}\left(\displaystyle\frac{1-x}{1+x}\right)=dxdtan−1(1+x1−x)= ____________.
x=a(θ-sinθ), y=a(1-cosθ) হলে -
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