Given-a-function-R-R-with-2-x-2-Find-the-value-of-d-2-dx-2-when-8-

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

Given a function ψ : R \to R

with ψ (θ) = θ^{2} - x^{2} . Find the value

of \frac{d^{2} ψ (θ)}{{dx}^{2}} when θ = 8

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

θ independent variable ?

Commented by bemath last updated on 13/Oct/20

no sir

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

\frac{d}{d x} ψ = 2 θ \frac{d}{d x} θ - 2 x

\frac{d^{2}}{{d x}^{2}} ψ = 2 {(\frac{d θ}{d x})}^{2} + 2 θ \frac{d^{2} θ}{{d x}^{2}} - 2

i t h i n k i f θ i s n o t i n d e p e n d e n t o f

y i u w i l l n e e d t o k n o w θ (x)

Commented by bemath last updated on 13/Oct/20

sir in line 2 it should be

\frac{d^{2} ψ}{{dx}^{2}} = 2 {(\frac{d θ}{dx})}^{2} + 2 θ \frac{d^{2} θ}{{dx}^{2}} - 2 ?

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

yes . corrected in red

Commented by Dwaipayan Shikari last updated on 13/Oct/20

B u t i n t h e q u e s t i o n θ i s i n d e p e n d e n t o f x

ψ (θ) = θ^{2} - x^{2} (A n d ψ (θ) = θ^{2} - x^{2} W h e r e - x^{2} i s c o n s t a n t)

ψ^{'} (θ) = 2 θ \frac{d θ}{d x} - 2 x

ψ^{″} (θ) = 2 \frac{d θ}{d x} + 2 θ \frac{d^{2} θ}{{d x}^{2}} - 2 \frac{d θ}{d x} = 0

ψ^{″} (θ) = - 2

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

It was already replied that θ is not

independent variable .