<s>[INST] 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
 [/INST] 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 $y'(2)$.
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