পর্যায়ক্রমিক অন্তরজ (Successive Differentiation)
y=lnex2 y=\operatorname{ln} e^{x^{2}} y=lnex2 হলে y2=? y_{2}=? y2=?
y=lnex2=x2lne=x2 \begin{aligned} y & =\ln e^{x^{2}} \\ & =x^{2} \ln e \\ & =x^{2}\end{aligned} y=lnex2=x2lne=x2
⇒y1=2x∴y2=2 \begin{array}{l}\Rightarrow y_{1}=2 x \\ \therefore y_{2}=2\end{array} ⇒y1=2x∴y2=2
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
