Question Number 117704 by bemath last updated on 13/Oct/20
$$\mathrm{Given}\:\mathrm{a}\:\mathrm{function}\:\psi:\mathbb{R}\rightarrow\mathbb{R} \\ $$$$\mathrm{with}\:\psi\left(\theta\right)\:=\:\theta^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} .\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:\frac{\mathrm{d}^{\mathrm{2}} \psi\left(\theta\right)}{\mathrm{dx}^{\mathrm{2}} }\:\:\mathrm{when}\:\theta=\mathrm{8} \\ $$
Commented by prakash jain last updated on 13/Oct/20
$$\theta\:\mathrm{independent}\:\mathrm{variable}? \\ $$
Commented by bemath last updated on 13/Oct/20
$$\mathrm{no}\:\mathrm{sir}\: \\ $$
Commented by prakash jain last updated on 13/Oct/20
$$\frac{{d}}{{dx}}\psi=\mathrm{2}\theta\frac{{d}}{{dx}}\theta−\mathrm{2}{x} \\ $$$$\frac{{d}^{\mathrm{2}} }{{dx}^{\mathrm{2}} }\psi=\mathrm{2}\left(\frac{{d}\theta}{{dx}}\right)^{\mathrm{2}} +\mathrm{2}\theta\frac{{d}^{\mathrm{2}} \theta}{{dx}^{\mathrm{2}} }−\mathrm{2} \\ $$$${i}\:{think}\:{if}\:\theta\:{is}\:{not}\:{independent}\:{of} \\ $$$${yiu}\:{will}\:{need}\:{to}\:{know}\:\theta\left({x}\right) \\ $$
Commented by bemath last updated on 13/Oct/20
$$\mathrm{sir}\:\mathrm{in}\:\mathrm{line}\:\mathrm{2}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\: \\ $$$$\frac{\mathrm{d}^{\mathrm{2}} \psi}{\mathrm{dx}^{\mathrm{2}} }\:=\:\mathrm{2}\left(\frac{\mathrm{d}\theta}{\mathrm{dx}}\right)^{\mathrm{2}} +\mathrm{2}\theta\:\frac{\mathrm{d}^{\mathrm{2}} \theta}{\mathrm{dx}^{\mathrm{2}} }\:−\mathrm{2}\:? \\ $$
Commented by prakash jain last updated on 13/Oct/20
$$\mathrm{yes}.\:\mathrm{corrected}\:\mathrm{in}\:\mathrm{red} \\ $$
Commented by Dwaipayan Shikari last updated on 13/Oct/20
$${But}\:{in}\:{the}\:{question}\:\theta\:{is}\:{independent}\:{of}\:{x} \\ $$$$\psi\left(\theta\right)=\theta^{\mathrm{2}} −{x}^{\mathrm{2}} \:\:\:\:\:\:\:\:\left({And}\:\psi\left(\theta\right)=\theta^{\mathrm{2}} −{x}^{\mathrm{2}} \:\:\:\:\:\:{Where}\:−{x}^{\mathrm{2}} \:{is}\:\:{constant}\right) \\ $$$$\psi'\left(\theta\right)=\mathrm{2}\theta\frac{{d}\theta}{{dx}}−\mathrm{2}{x} \\ $$$$\psi''\left(\theta\right)=\mathrm{2}\frac{{d}\theta}{{dx}}+\mathrm{2}\theta\frac{{d}^{\mathrm{2}} \theta}{{dx}^{\mathrm{2}} }−\mathrm{2}\:\:\:\:\:\:\:\frac{{d}\theta}{{dx}}=\mathrm{0} \\ $$$$\psi''\left(\theta\right)=−\mathrm{2} \\ $$
Commented by prakash jain last updated on 13/Oct/20
$$\mathrm{It}\:\mathrm{was}\:\mathrm{already}\:\mathrm{replied}\:\mathrm{that}\:\theta\:\mathrm{is}\:\mathrm{not} \\ $$$$\mathrm{independent}\:\mathrm{variable}. \\ $$