# What is the derivative of f(t) = (t^2-1 , te^(2t-1) ) ?

Apr 24, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{2 t - 1} \left(1 + 2 t\right)}{2 t}$

#### Explanation:

The derivative of a parametric function is defined by:
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$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$
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Here $x = {t}^{2} - 1 \text{ }$and$\text{ } y = t {e}^{2 t - 1}$
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$\frac{\mathrm{dx}}{\mathrm{dt}} = 2 t$
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$\frac{\mathrm{dy}}{\mathrm{dt}} = {e}^{2 t - 1} + 2 t {e}^{2 t - 1} = {e}^{2 t - 1} \left(1 + 2 t\right)$
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$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}} = \frac{{e}^{2 t - 1} \left(1 + 2 t\right)}{2 t}$