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

Nov 21, 2017

See explanation.

#### Explanation:

It depends on what derivative you want. If you want $\frac{\mathrm{dy}}{\mathrm{dt}} \mathmr{and} \frac{\mathrm{dx}}{\mathrm{dt}}$, simply differentiate the x and y functions with respect to t, the same way you would differentiate a function f(x) with respect to x.

$x \left(t\right) = {e}^{1 - t} \to \frac{\mathrm{dx}}{\mathrm{dt}} = - {e}^{1 - t}$
y(t) = tan^2(t) = (tan t)(tan t) -> dy/dt = 2tan(t)d/dt(tan t) = (2 tan t)/(sec^2(t)

If you specifically want $\frac{\mathrm{dy}}{\mathrm{dx}}$, we must use the chain rule .

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

This last term is simply one over $\frac{\mathrm{dx}}{\mathrm{dt}}$...

dy/dx = (2tant)/(sec^2(t)) * 1/(-e^(1-t)

Recalling that ${e}^{a - b} = {e}^{a} / {e}^{b} \ldots$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \tan t}{{\sec}^{2} \left(t\right)} \cdot - \frac{1}{\left(\frac{e}{e} ^ t\right)} = \frac{2 \tan t}{{\sec}^{2} \left(t\right)} \cdot - {e}^{t} / e = \frac{2 \tan t}{{\sec}^{2} \left(t\right)} \cdot - {e}^{t - 1} = \frac{\left(- 2 \tan t\right) \left({e}^{t - 1}\right)}{{\sec}^{2} \left(t\right)}$