বিপরীত ফাংশন ও পরামিতিক ফাংশনের অন্তরজ
tan−14x1−4x \tan ^{-1} \frac{4 \sqrt{x}}{1-4 x} tan−11−4x4x xxx এর সাপেক্ষে অন্তরক সহগ নিচের কোনটি?
2x(1−4x)\frac{2}{\sqrt{x}(1-4 x)}x(1−4x)2
−2x(1+4x)\frac{-2}{\sqrt{x}(1+4 x)}x(1+4x)−2
4x(1+4x)\frac{4}{\sqrt{x}(1+4 x)}x(1+4x)4
2x(1+4x)\frac{2}{\sqrt{x}(1+4 x)}x(1+4x)2
Solve:=tan−12⋅2x1−(2x)2=2tan−1(2x)[∵tan−12x1−x2=2tan−1x] \begin{array}{c} =\tan ^{-1} \frac{2 \cdot 2 \sqrt{x}}{1-(2 \sqrt{x})^{2}}=2 \tan ^{-1}(2 \sqrt{x}) \\ {\left[\because \tan ^{-1} \frac{2 x}{1-x^{2}}=2 \tan ^{-1} x\right]} \end{array} =tan−11−(2x)22⋅2x=2tan−1(2x)[∵tan−11−x22x=2tan−1x]
∴ddx(tan−14x1−4x)=ddx{2tan−1(2x)}=211+(2x)2ddx(2x)=21+4x⋅2⋅12x=2x(1+4x) (Ans.) \begin{aligned} \therefore & \frac{d}{d x}\left(\tan ^{-1} \frac{4 \sqrt{x}}{1-4 x}\right)=\frac{d}{d x}\left\{2 \tan ^{-1}(2 \sqrt{x})\right\} \\ & =2 \frac{1}{1+(2 \sqrt{x})^{2}} \frac{d}{d x}(2 \sqrt{x}) \\ & =\frac{2}{1+4 x} \cdot 2 \cdot \frac{1}{2 \sqrt{x}}=\frac{2}{\sqrt{x}(1+4 x)} \text { (Ans.) } \end{aligned} ∴dxd(tan−11−4x4x)=dxd{2tan−1(2x)}=21+(2x)21dxd(2x)=1+4x2⋅2⋅2x1=x(1+4x)2 (Ans.)
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 ?
xxx এর সাপেক্ষে অন্তরক সহগ নিচের কোনটি? sin−1x5 \sqrt{\sin ^{-1} x^{5}} sin−1x5
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 -
নিচের কোনটি সঠিক?
