Question

${(({ \frac{1 + {t}^{2} }{1 - {t}^{2} }) }^{2} - 1) }^{ \frac{1}{2} }$
Please find the derivative of the above function..... ​

Answer 1

Answer:

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Answer 2

Answer:

$\frac{1}{ \sqrt{( \frac{1 + {t}^{2} }{1 - {t}^{2} } ) ^{2} - 1} } \times \frac{1 + {t}^{2} }{1 - {t}^{2} } \times \frac{ - 4 {t}^{3} }{1 + {t}^{4} - 2 {t}^{2} }$

Step-by-step explanation:

we know that

${n}^{ \frac{1}{2} } = \sqrt{n}$

therefore,

$(( \frac{ {1 + {t}^{2} } }{1 - {t}^{2} } )^{2} - 1) ^{ \frac{1}{2} }$

$= \sqrt{( \frac{1 + {t}^{2} }{1 - {t}^{2} }) ^{2} - 1}$

Now diffrrentiate with respect to t

$\frac{dy}{dt} = \frac{1}{2 \sqrt{( \frac{1 + {t}^{2} }{1 - {t}^{2} } ) ^{2} - 1 } } \times 2 \frac{1 + {t}^{2} }{1 - {t}^{2} } \times \frac{2t - 2 {t}^{3} - 2t - 2 {t}^{3} }{1 + {t}^{4} - 2 {t}^{2} }$

$= \frac{1}{ \sqrt{( \frac{1 + {t}^{2} }{1 - {t}^{2} }) ^{2} - 1 } } \times \frac{1 + {t}^{2} }{1 - {t}^{2} } \times \frac{ - 4 {t}^{3} }{1 + {t}^{4} - 2 {t}^{2} }$