পর্যায়ক্রমিক অন্তরজ (Successive Differentiation)
y=2xln11−x;f(x)=(1−x2)y2−xy2−a2y y=2^{x} \ln \frac{1}{1-x} ; f(x)=\left(1-x^{2}\right) y_{2}-x y_{2}-a^{2} y y=2xln1−x1;f(x)=(1−x2)y2−xy2−a2y
দেখাও যে, limx→0x1−1+x=−2 \lim _{x \rightarrow 0} \frac{x}{1-\sqrt{1+x}}=-2 limx→01−1+xx=−2
x x x এর সাপেক্ষে dydx \frac{d y}{d x} dxdy নির্ণয় কর।
দেখাও যে, f(x)=0 f(x)=0 f(x)=0, যখন sin−1x=lnya \sin ^{-1} x=\frac{\ln y}{a} sin−1x=alny.
f(x)=lnx,g(x)=(x+1+x2)f(x)=\ln x, g(x)=\left(x+\sqrt{1+x^{2}}\right)f(x)=lnx,g(x)=(x+1+x2)
দৃশ্যকল্প: f(x,y)=x+y−2,t=2sin−1x\mathrm{ f(x, y)=\sqrt{x}+\sqrt{y}-\sqrt{2}}, \mathrm{t=2 \sin ^{-1} x} f(x,y)=x+y−2,t=2sin−1x.
দৃশ্যকল্প: g(x)=xcosec−11x,y=cos(msin−1p) \mathrm{g}(\mathrm{x})=\mathrm{xcosec}^{-1} \frac{1}{\mathrm{x}}, \mathrm{y}=\cos \left(\mathrm{msin}^{-1} \mathrm{p}\right) g(x)=xcosec−1x1,y=cos(msin−1p)
y=eθg(x) \mathrm{y}=\mathrm{e}^{\operatorname{\theta g}(\mathrm{x})} y=eθg(x) এবং f(x)=1x \mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}} f(x)=x1
