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Question Number 117704 by bemath last updated on 13/Oct/20

$$\mathrm{Given}\mathrm{a}\mathrm{function}\psi :\mathbb{R}\to \mathbb{R}$$$$\mathrm{with}\psi \left(\theta \right)={\theta }^2-{\mathrm{x}}^2.\mathrm{Find}\mathrm{the}\mathrm{value}$$$$\mathrm{of}\frac{{\mathrm{d}}^2\psi \left(\theta \right)}{{\mathrm{dx}}^2}\mathrm{when}\theta =8$$

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

$$\frac{d}{dx}\psi =2\theta \frac{d}{dx}\theta -2x$$$$\frac{d^2}{{dx}^2}\psi =2{\left(\frac{d\theta }{dx}\right)}^2+2\theta \frac{d^2\theta }{{dx}^2}-2$$$$ithinkif\theta isnotindependentof$$$$yiuwillneedtoknow\theta \left(x\right)$$

Commented by bemath last updated on 13/Oct/20

$$\mathrm{sir}\mathrm{in}\mathrm{line}2\mathrm{it}\mathrm{should}\mathrm{be}$$$$\frac{{\mathrm{d}}^2\psi }{{\mathrm{dx}}^2}=2{\left(\frac{\mathrm{d}\theta }{\mathrm{dx}}\right)}^2+2\theta \frac{{\mathrm{d}}^2\theta }{{\mathrm{dx}}^2}-2?$$

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

$$Butinthequestion\theta isindependentofx$$$$\psi \left(\theta \right)={\theta }^2-x^2(And\psi \left(\theta \right)={\theta }^2-x^{2}Where-x^{2}isconstant)$$$${\psi }^{\prime }\left(\theta \right)=2\theta \frac{d\theta }{dx}-2x$$$${\psi }^{″}\left(\theta \right)=2\frac{d\theta }{dx}+2\theta \frac{d^2\theta }{{dx}^2}-2\frac{d\theta }{dx}=0$$$${\psi }^{″}\left(\theta \right)=-2$$

Commented by prakash jain last updated on 13/Oct/20

$$\mathrm{It}\mathrm{was}\mathrm{already}\mathrm{replied}\mathrm{that}\theta \mathrm{is}\mathrm{not}$$$$\mathrm{independent}\mathrm{variable}.$$

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