Question

X = e^t/2 + e^-1/2 and y= e^t - e^-1/2 find dy/dx teachoo

Answer 1

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$= > x = {e}^{ \frac{t}{2} } + {e}^{ - \frac{1}{2} } \\ \\ = > y = {e}^{t} - {e}^{ - \frac{1}{2} } \\ \\ \frac{dx}{dt} = \frac{d( {e}^{ \frac{t}{2} }) }{dt} + \frac{d( {e}^{ - \frac{1}{2} } )}{dt} \\ \\ \: \: \: \: \: \: \: = \frac{t}{2} {e}^{ \frac{t}{2} - 1 } = \frac{t {e}^{ \frac{t}{2} } }{2e} = \frac{t \sqrt{ {e}^{t} } }{2e} \\ \\ \frac{dy}{dt} = \frac{d ({e}^{t}) }{dt} - \frac{d( {e}^{ - \frac{1}{2} }) }{dt} \\ \\ \: \: \: \: \: \: \: = t {e}^{t - 1} = \frac{t {e}^{t} }{e} \\ \\ now.... \\ \\ \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = \frac{t {e}^{t} }{e} \times \frac{2e}{t \sqrt{ {e}^{t} } } \\ \\ \: \: \: \: \: \: \: = 2 \sqrt{ {e}^{t} } = 2 {e}^{ \frac{t}{2} }$
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