# What is the slope of the curve at t=3 assuming that the equations define x and y implicitly as differentiable functions x=f(t), y=g(t), and x=t^5+t, y+4t^5=4x+t^4?

Mar 10, 2015

If:
$x = {t}^{5} + t$
$y + 4 {t}^{5} = 4 \left({t}^{5} + t\right) + {t}^{4}$
$y = - 4 {t}^{5} + 4 {t}^{5} + 4 t + {t}^{4} = {t}^{4} + 4 t$

Now:
$\frac{\mathrm{dx}}{\mathrm{dt}} = 5 {t}^{4} + 1$
$\frac{\mathrm{dy}}{\mathrm{dt}} = 4 {t}^{3} + 4$

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

So:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 {t}^{3} + 4}{5 {t}^{4} + 1}$

For $t = 3$ the slope will be:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 \cdot \left(27 + 4\right)}{5 \cdot \left(81 + 1\right)} = \frac{124}{410} = \frac{62}{205} = 3.31$