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
On
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 $$ y + 22 = -28(x - 2) \text{.} $$
On
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.
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 $y'(2)$.