Question

find the derivative of given function w.r.t independent variable 't' :
f(t)= t²-1/ t²+t-2​​

Answer 1

$\huge{\underline{\mathtt{\red{A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}}}}$

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

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

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

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

$f(t) = \frac{(t + 1)}{(t + 2)}$

By using division rule:

$\frac{d \: (t + 1)}{dt(t + 2)}$

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

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

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

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

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

Thank you!!

Answer 2

✫$\LARGE{\mathfrak{\underline{\underline{\ Solution}}}}$✫

$: \implies \bf \large{ \frac{d}{dx} ( \frac{ {t}^{2} - 1 }{ {t}^{2} + t - 2} )}$

$: \implies \bf \large{ \frac{d}{dx} ( \frac{(t + 1)(t - 1)}{t(t + 2) - 1(t + 2)} )}$

$: \implies \bf \large{ \frac{d}{dx} ( \frac{(t + 1) \cancel{(t - 1)}}{ \cancel{(t - 1)}(t + 2)} ) }$

$: \implies \bf \large{ \frac{d}{dx}( \frac{t + 1}{t + 2} )}$

$: \implies \bf \large{ \frac{(t + 2) \frac{d}{dx}(t + 1) - (t + 1) \frac{d}{dx} (t + 2) }{ {(t + 2)}^{2} } }$

$: \implies \bf \large{ \frac{(t + 2) \times 1 - (t + 1) \times 1}{ {(t + 2)}^{2} } }$

$: \implies \bf \large{ \frac{t + 2 - t - 1}{ {(t + 2)}^{2} } }$

$\begin{lgathered}\therefore \underline{ \boxed{\frak{ \dfrac{1}{ (t + 2) ^{2} }}}} \\\end{lgathered}$