Question

If x=1/2(t+1/t) y=1/2(t-1/t) then find dy/dx

Answer 1

Answer:

$\dfrac{dy}{dx} \: = \dfrac{ {t}^{2} + 1}{ {t}^{2} - 1 }$

Step-by-step explanation:

Given--->

$x \: = \dfrac{1}{2} (t \: + \dfrac{1}{t} )$

$and$

$y \: = \dfrac{1}{2} (t \: - \dfrac{1}{t} )$

To find ---->

$value \: of \: \dfrac{dy}{dx}$

Concept used ---->

1)

$\dfrac{dy}{dx} = \frac{ \dfrac{dy}{dt} }{ \dfrac{dx}{dt} }$

2)

$\dfrac{d}{dx} ( \: \dfrac{u}{v} \: \: ) \: = \dfrac{v \: \dfrac{du}{dx} \: \: - \: u \: \dfrac{dv}{dx} }{ {v}^{2} }$

3)

$\dfrac{d}{dx} ( \: {x}^{n} ) = \: n \: {x}^{n - 1}$

Solution---->

$now$

$x \: = \frac{1}{2} ( \: t \: + \: \dfrac{1}{t} \: )$

$x \: = \dfrac{1}{2} \: ( \: \dfrac{ {t}^{2} \: + \: 1}{t} \: )$

$x \: = \: \dfrac{ {t}^{2} \: + \: 1 }{2t}$

$differentiating \: with \: respect \: to \: x \: we \: get$

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

$\dfrac{dx}{dt} \: = \dfrac{2t \: ( \: 2t \: + \: 0 \: ) \: - \: ( {t}^{2} \: + 1 \: ) \: ( \: 2 \: ) }{4 {t}^{2} }$

$\dfrac{dx}{dt} \: = \dfrac{4 {t}^{2} \: - 2 {t}^{2} \: - \: 2 }{4 {t}^{2} }$

$\dfrac{dx}{dt} \: = \dfrac{2 {t}^{2} \: - \: 2 }{4 {t}^{2} }$

$\dfrac{dx}{dt} \: = \: \dfrac{ {t}^{2} \: - \: 1}{2 {t}^{2} }$

$now$

$y \: = \dfrac{1}{2} \: ( \: t \: - \dfrac{1}{t} )$

$y \: = \: \dfrac{ {t}^{2} \: - \: 1 }{2 \: t}$

$differentiting \: with \: respect \: to \: t \: we \: get$

$\dfrac{dy}{dt} \: = \dfrac{2t \: \dfrac{d}{dt} \: ( {t}^{2} \: - \: 1) \: - ( {t}^{2} \: - 1) \: \dfrac{d}{dt} \: (2t) }{ {(2t)}^{2} }$

$\dfrac{dy}{dt} \: = \dfrac{2t \: (2t \: - \: 0) \: - ( {t}^{2} \: - 1) \: (2) }{4 {t}^{2} }$

$\dfrac{dy}{dt} \: = \dfrac{4 {t}^{2} \: - 2 {t}^{2} \: + 2 }{4 {t}^{2} }$

$\dfrac{dy}{dt} \: = 2 \: \dfrac{ ({t}^{2} \: + 1) }{4 {t}^{2} }$

$\dfrac{dy}{dt} \: = \dfrac{ {t}^{2} \: + 1}{2 {t}^{2} }$

$now$

$\dfrac{dy}{dx} \: = \dfrac{ \dfrac{dy}{dt} }{ \dfrac{dx}{dt} }$

$\dfrac{dy}{dx} \: = \dfrac{ \dfrac{ {t}^{2} \: + \: 1 }{2 {t}^{2} } }{ \dfrac{ {t}^{2} \: - 1 }{2 {t}^{2} } }$

$\dfrac{dy}{dx} \: = \dfrac{ {t}^{2} \: + \: 1 }{ {t}^{2} \: - 1 }$