Answer
The slope of the tangent line of the curve at $P$ is $-3$
The equation of the tangent line of the curve at $P$ is $y=-3x+4$
Work Step by Step
$$y=2-x^3\hspace{1cm} P(1,1)$$
1) First, take $Q(1+h,y)$ to be a nearby point of $P$ on the graph of the function.
$$y=2-(1+h)^3=2-(1+3h+3h^2+h^3)$$ $$y=2-1-3h-3h^2-h^3=1-3h-3h^2-h^3$$
So $Q(1+h,1-3h-3h^2-h^3)$
2) Find the slope of the secant $PQ$:
$$\frac{\Delta y}{\Delta x}=\frac{1-3h-3h^2-h^3-1}{1+h-1}=\frac{-3h-3h^2-h^3}{h}=-3-3h-h^2$$
3) Find out what happens if $Q$ approaches $P$
As $Q$ approaches $P$, $1+h$ will gradually approach $1$, while $1-3h-3h^2-h^3$ also approaches $1$. Both of these mean that $h$ will approach $0$ and hence secant slope $-3-3h-h^2$ will approach $-3$.
So we take $-3$ to be the slope of the tangent line of the curve at $P$.
4) The equation of the tangent line at $P$ would have this form: $$y=-3x+b$$
Replace the coordinates of $P$ here to find $b$:
$$(-3)\times1+b=1$$ $$-3+b=1$$ $$b=4$$
Therefore, the equation of the tangent line of the curve at $P$ is $y=-3x+4$
