Plzzz solve my doubt!!!!!????

  Dear Student
$$\begin{gathered} r=\frac{12}{\theta }-\frac{4}{{\theta }^3}+\frac{1}{{\theta }^4} \\ \frac{dr}{d\theta }=\frac{d}{d\theta }\left(12{\theta }^{-1}-4{\theta }^{-3}+{\theta }^{-4}\right) \\ \frac{dr}{d\theta }=\left(-12{\theta }^{-2}-(-3)4{\theta }^{-4}+(-4){\theta }^{-5}\right) \\ \frac{dr}{d\theta }=\left(-12{\theta }^{-2}+12{\theta }^{-4}-4{\theta }^{-5}\right) \\ \frac{dr}{d\theta }=\left(-\frac{12}{{\theta }^2}+\frac{12}{{\theta }^4}-\frac{4}{{\theta }^5}\right) ...\left(1\right) \\ \frac{d^{2}r}{d{\theta }^2}=\frac{d}{d\theta }\left(-12{\theta }^{-2}+12{\theta }^{-4}-4{\theta }^{-5}\right) \\ \frac{d^{2}r}{d{\theta }^2}=\left(-12(-2){\theta }^{-3}+12(-4){\theta }^{-5}-4(-5){\theta }^{-6}\right) \\ \frac{d^{2}r}{d{\theta }^2}=\left(24{\theta }^{-3}-48{\theta }^{-5}+20{\theta }^{-6}\right) \\ \frac{d^{2}r}{d{\theta }^2}=\left(\frac{24}{{\theta }^3}-\frac{48}{{\theta }^5}+\frac{20}{{\theta }^6}\right) ...\left(2\right) \end{gathered}$$
Equation 1 and 2 are required expressions

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