Question

guys please...fast...​

Answer 1

Answer:

Given :

A function of x wrt t has been provided as follows :

$x = \sqrt{t} + \dfrac{1}{ \sqrt{t} }$

To find:

Derivative of x with respect to t

Calculation:

$x = \sqrt{t} + \dfrac{1}{ \sqrt{t} }$

$= > \dfrac{dx}{dt} = \dfrac{d( \sqrt{t} )}{dt} + \dfrac{d( {t}^{ - \frac{1}{2} }) }{dt}$

$= > \dfrac{dx}{dt} = \dfrac{d( {t}^{ \frac{1}{2} } )}{dt} + \dfrac{d( {t}^{ - \frac{1}{2} }) }{dt}$

$= > \dfrac{dx}{dt} = \dfrac{1}{2} {t}^{ (\frac{1}{2} - 1) } - \dfrac{1}{2} {t}^{ (- \frac{1}{2} - 1)}$

$= > \dfrac{dx}{dt} = \dfrac{1}{2} {t}^{ ( - \frac{1}{2} ) } - \dfrac{1}{2} {t}^{ (- \frac{3}{2} )}$

$= > \dfrac{dx}{dt} = \dfrac{1}{2 \sqrt{t} } - \dfrac{1}{2t \sqrt{t} }$

So final answer :

$\boxed{ \red{ \bold{ \dfrac{dx}{dt} = \dfrac{1}{2 \sqrt{t} } - \dfrac{1}{2t \sqrt{t} } }}}$

Answer 2

Solution :

✴ Given :-

✏ A function of x in terms of time t

$\boxed{\sf{\pink{x=\sqrt{t}+\dfrac{1}{\sqrt{t}}}}}$

✴ To Find :-

✏ dx/dt = ?

✴ Calculation :-

$\dashrightarrow\sf\:\dfrac{dx}{dt}=\dfrac{d(\sqrt{t}+\frac{1}{\sqrt{t}})}{dt}\\ \\ \dashrightarrow\sf\:\dfrac{dx}{dt}=\dfrac{d(t^{\frac{1}{2}}+t^{-\frac{1}{2}})}{dt}\\ \\ \dashrightarrow\sf\:\dfrac{dx}{dt}=\dfrac{1}{2}t^{\frac{1}{2}-1}-\frac{1}{2}t^{-\frac{1}{2}-1}\\ \\ \dashrightarrow\sf\:\dfrac{dx}{dt}=\dfrac{1}{2}t^{-\frac{1}{2}}-\dfrac{1}{2}t^{-\frac{3}{2}}\\ \\ \dashrightarrow\sf\:\dfrac{dx}{dt}=\dfrac{1}{2\sqrt{t}}-\dfrac{1}{2\sqrt{t^3}}\\ \\ \dashrightarrow\:\boxed{\tt{\purple{\large{\dfrac{dx}{dt}=\dfrac{1}{2\sqrt{t}}-\dfrac{1}{2t\sqrt{t}}}}}}$