বিপরীত ফাংশন ও পরামিতিক ফাংশনের অন্তরজ
যদি y= sin-1x হয়, তবে y1/y2 এর মান কোনটি?
None
y=sin−1x⇒y1=11−x2y2=1−x2×0−1121−x2×(−2x)(1−x2)=x(1−x2)(1−x2)∴y1y2=11−x2×(1−x2)(1−x2)x=1−x2x=1x−x2x=1x−x \begin{array}{c}y=\sin ^{-1} x \Rightarrow y_{1}=\frac{1}{\sqrt{1-x^{2}}} \\ y_{2}=\frac{\sqrt{1-x^{2}} \times 0-1 \frac{1}{2 \sqrt{1-x^{2}}} \times(-2 x)}{\left(1-x^{2}\right)}=\frac{x}{\left(1-x^{2}\right)\left(\sqrt{1-x^{2}}\right)} \\ \therefore \frac{y_{1}}{y_{2}}=\frac{1}{\sqrt{1-x^{2}}} \times \frac{\left(1-x^{2}\right)\left(\sqrt{1-x^{2}}\right)}{x}=\frac{1-x^{2}}{x}=\frac{1}{x}-\frac{x^{2}}{x}=\frac{1}{x}-x\end{array} y=sin−1x⇒y1=1−x21y2=(1−x2)1−x2×0−121−x21×(−2x)=(1−x2)(1−x2)x∴y2y1=1−x21×x(1−x2)(1−x2)=x1−x2=x1−xx2=x1−x
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 ?
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)= ____________.
