Find the slope of the tangent line to the graph of the given function at the given value of x. Find the slope of the tangent line to the graph of the given function at the given value of x. Find the equation of the tangent line. : $y=x^4-5x^3+2; x=2$
I understand that the slope of the line follows the equation of
$$\lim_{h\to 0} \frac{f(a+h)-f(a)}h.$$
But I don't know how to go about the rest
 A: The easiest equation is point-slope form because (1) as you observe, the derivative gives this slope and (2) it is very easy to get the $y$-coordinate of a point on the curve given its $x$-coordinate.  So, the line through the point $(x_0,y_0)$ of slope $m$ has the equation
$$  y - y_0 = m(x - x_0)  \text{.}  $$
The above should be enough for you to finish your problem.  If you need more...

 $\mathrm{d}y/\mathrm{d}x = 4x^3 - 15x^2$.  At x = 2, this evaluates to $4 \cdot 8 - 15 \cdot 4 = -28$, so the slope of the line in question is $-28$.
 Then, evaluating for $y$ when $x = x_0 = 2$, we have $y_0 = x_0^4 - 5 x_0^3 + 2 = 16 - 5 \cdot 8 + 2 = -22 \text{.}$

Finally, should you still not see how to assemble these three numbers into the equation...

 Then an equation of the line is $$  y - (-22) = -28(x - 2)  \text{,}  $$ although it is easy to clean this up a little, to $$  y + 22 = -28(x - 2)  \text{.}  $$

A: The slope should be the value of the derivative y(x)' at point (2,y(2)), where y(2)=-22.
Read How to get derivatives of polynomials for the details , you will learn  ($x^n$)'=n($x^{n-1}$) .
Given  y=$x^4$-5$x^3$+2, $y(x)'$ = $4x^3$-15$x^2$,   y(2)'=-28
From point-slope formula,  the line equation is  y+22=(x-2)(-28),  or   y=-28x+34.
A: If you use the h-method you start with
$$\lim_{h \rightarrow 0} \frac{y(a+h)-y(a)}{h}=\lim_{h \rightarrow 0} \frac{(a+h)^4-5\cdot (a+h)^3+2-(a^4-5\cdot a^3+2)}{h}$$
To resolve  $(a+h)^4$ you can use the binomial theorem: $(x+y)^n=\sum\limits_{k=0}^n\binom{n}x\cdot x^k\cdot y^{n-k}$
$$=\lim_{h \rightarrow 0}\frac{a^4+4a^3h+6a^2h^2+4ah^3+h^4-5a^3-15a^2h-15ah^2-5h^3+2-a^4+5\cdot a^3-2}{h}$$
Next we eliminate the terms with opposite signs, e.g. $a^4$ and $-a^4$
$$=\lim_{h \rightarrow 0}\frac{4a^3h+6a^2h^2+4ah^3+h^4-15a^2h-15ah^2-5h^3}{h}$$
Factoring out $h$ at the numerator.
$$=\lim_{h \rightarrow 0}\frac{h\cdot \left(4a^3+6a^2h+4ah^2+h^3-15a^2-15ah-5h^2\right)}{h}$$
Cancelling $h$
$$=\lim_{h \rightarrow 0}4a^3+6a^2h+4ah^2+h^3-15a^2-15ah-5h^2$$
$$=4a^3-15a^2=y'(a)$$
This is the derivative of $y(x)$ at value $x=a$ and therefore the slope. Finally you can insert $a=2$ to obtain $y'(2)$.
