# What is the derivative of f(t) = (e^t/t +e^t, e^t-cost ) ?

Jun 14, 2017

$f ' \left(t\right) = \left\langle{e}^{t} \left(1 + \frac{1}{t} - \frac{1}{t} ^ 2\right) , {e}^{t} + \sin t\right\rangle$

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

#### Explanation:

There are two different derivatives depending on what you want to know about the function. One of them provides the velocity (a vector), and one provides the slope (a scalar).

The velocity vector for a function at time $t$ is found by:

$f ' \left(x\right) = \left\langle\frac{\mathrm{dx}}{\mathrm{dt}} , \frac{\mathrm{dy}}{\mathrm{dt}}\right\rangle$

So we need to find $\frac{\mathrm{dx}}{\mathrm{dt}}$ and $\frac{\mathrm{dy}}{\mathrm{dt}}$. They are:

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

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

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left({e}^{t} - \cos t\right) = {e}^{t} + \sin t$

Therefore, our velocity derivative is:

$f ' \left(x\right) = \left\langle{e}^{t} \left(1 + \frac{1}{t} - \frac{1}{t} ^ 2\right) , {e}^{t} + \sin t\right\rangle$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The slope of this function can be found with:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\mathrm{dy} \text{/"dt)/(dx"/} \mathrm{dt}\right)$

Since we already found $\frac{\mathrm{dy}}{\mathrm{dt}}$ and $\frac{\mathrm{dx}}{\mathrm{dt}}$ above, we can use those same functions here to shorten the explanation:

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

To make this look a bit simpler, we can multiply the top and bottom by ${t}^{2}$ to eliminate the fractions on the bottom.

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