বিপরীত ফাংশন ও পরামিতিক ফাংশনের অন্তরজ
dydx \frac{d y}{d x} dxdy নিচের কোনটি? x=t,y=t−1t x=\sqrt{t}, y=t-\frac{1}{\sqrt{t}} x=t,y=t−t1
2t+12t 2 \sqrt{t}+\frac{1}{2t}2t+2t1
2t−1t 2 \sqrt{t}-\frac{1}{t}2t−t1
4t+1t 4 \sqrt{t}+\frac{1}{t}4t+t1
2t+1t 2 \sqrt{t}+\frac{1}{t}2t+t1
Solve:dxdt=ddt(t)=12t and dydt=ddt(t−1t)=ddt(t−t−12)=1−(−12)t−12−1=1+12tt=12t(2t+1t)∴dydx=dydt×dtdx=12t(2t+1t)×2t1=2t+1t \begin{array}{l} \frac{d x}{d t}=\frac{d}{d t}(\sqrt{t})=\frac{1}{2 \sqrt{t}} \text { and } \\ \frac{d y}{d t}=\frac{d}{d t}\left(t-\frac{1}{\sqrt{t}}\right)=\frac{d}{d t}\left(t-t^{-\frac{1}{2}}\right) \\ =1-\left(-\frac{1}{2}\right) t^{-\frac{1}{2}-1}=1+\frac{1}{2 t \sqrt{t}} \\ = \frac{1}{2 \sqrt{t}}\left(2 \sqrt{t}+\frac{1}{t}\right) \\ \therefore \quad \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{1}{2 \sqrt{t}}\left(2 \sqrt{t}+\frac{1}{t}\right) \times \frac{2 \sqrt{t}}{1} \\ =2 \sqrt{t}+\frac{1}{t} \end{array} dtdx=dtd(t)=2t1 and dtdy=dtd(t−t1)=dtd(t−t−21)=1−(−21)t−21−1=1+2tt1=2t1(2t+t1)∴dxdy=dtdy×dxdt=2t1(2t+t1)×12t=2t+t1
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)= ____________.
