# 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?

What is the slope of the curve at t=3 assuming that the equations define x and y implicitly as differentiable functions $x=f\left(t\right)$, $y=g\left(t\right)$, and $x={t}^{5}+t$, $y+4{t}^{5}=4x+{t}^{4}$?
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If:
$x={t}^{5}+t$
$y+4{t}^{5}=4\left({t}^{5}+t\right)+{t}^{4}$
$y=-4{t}^{5}+4{t}^{5}+4t+{t}^{4}={t}^{4}+4t$
Now
$\frac{dx}{dt}=5{t}^{4}+1$
$\frac{dy}{dt}=4{t}^{3}+4$
But
$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}$
So
$\frac{dy}{dx}=\frac{4{t}^{3}+4}{5{t}^{4}+1}$
For t=3 the slope will be:
$\frac{dy}{dx}=\frac{4\cdot \left(27+4\right)}{5\cdot \left(81+1\right)}=\frac{124}{410}=\frac{62}{205}=3.31$