${\rm{x}} = \frac{{{\rm{\alpha }}\left( {1 - {{\rm{t}}^2}} \right)}}{{1 + {{\rm{t}}^2}}},{\rm{\;y}} = \frac{{2{\rm{at}}}}{{1 + {{\rm{t}}^2}}}$What is $\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}}$ equal to?

Question

${\rm{x}} = \frac{{{\rm{\alpha }}\left( {1 - {{\rm{t}}^2}} \right)}}{{1 + {{\rm{t}}^2}}},{\rm{\;y}} = \frac{{2{\rm{at}}}}{{1 + {{\rm{t}}^2}}}$What is $\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}}$ equal to?

TuteeHUB · Accepted Answer

NA

TuteeHUB · Answer

$\frac{{{{\rm{a}}^2}}}{{{{\rm{y}}^2}}}$

TuteeHUB · Answer

$\frac{{{{\rm{a}}^2}}}{{{{\rm{x}}^2}}}$

TuteeHUB · Answer

$ - \frac{{{{\rm{a}}^2}}}{{{{\rm{x}}^2}}}$

TuteeHUB · Answer

$- \frac{{{{\rm{a}}^2}}}{{{{\rm{y}}^3}}}$

1.	\({\rm{x}} = \frac{{{\rm{\alpha }}\left( {1 - {{\rm{t}}^2}} \right)}}{{1 + {{\rm{t}}^2}}},{\rm{\;y}} = \frac{{2{\rm{at}}}}{{1 + {{\rm{t}}^2}}}\)What is \(\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}}\) equal to?
A.	\(\frac{{{{\rm{a}}^2}}}{{{{\rm{y}}^2}}}\)
B.	\(\frac{{{{\rm{a}}^2}}}{{{{\rm{x}}^2}}}\)
C.	\( - \frac{{{{\rm{a}}^2}}}{{{{\rm{x}}^2}}}\)
D.	\(- \frac{{{{\rm{a}}^2}}}{{{{\rm{y}}^3}}}\)
Answer» E.

\({\rm{x}} = \frac{{{\rm{\alpha }}\left( {1 - {{\rm{t}}^2}} \right)}}{{1 + {{\rm{t}}^2}}},{\rm{\;y}} = \frac{{2{\rm{at}}}}{{1 + {{\rm{t}}^2}}}\)What is \(\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}}\) equal to?

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