\(\displaystyle \begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=x + 9x^{2} - 6\text{ and we can find that }\\&f(x+h)=h + x + 9\cdot (h + x)^{2} - 6\\&f'(x)=\frac{h + x + 9\cdot (h + x)^{2} - 6-(x + 9x^{2} - 6)}{h}\\&f'(x)=\frac{h + 18\cdot hx + 9\cdot h^{2}}{h}\\&f'(x)=9\cdot h + 18x + 1\\&\text{Then, if we take the limit as h goes to zero, we find at }x=8:\\&145\end{align*}\)

\(\displaystyle \begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=2x + 6x^{2} + 10\text{ and we can find that }\\&f(x+h)=2\cdot h + 2x + 6\cdot (h + x)^{2} + 10\\&f'(x)=\frac{2\cdot h + 2x + 6\cdot (h + x)^{2} + 10-(2x + 6x^{2} + 10)}{h}\\&f'(x)=\frac{2\cdot h + 12\cdot hx + 6\cdot h^{2}}{h}\\&f'(x)=6\cdot h + 12x + 2\\&\text{Then, if we take the limit as h goes to zero, we find at }x=-10:\\&-118\end{align*}\)

\(\displaystyle \begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=- 10x - 6\text{ and we can find that }\\&f(x+h)=- 10\cdot h - 10x - 6\\&f'(x)=\frac{- 10\cdot h - 10x - 6-(- 10x - 6)}{h}\\&f'(x)=\frac{-10\cdot h}{h}\\&f'(x)=-10\\&\text{Then, if we take the limit as h goes to zero, we find at }x=-1:\\&-10\end{align*}\)

\(\displaystyle \begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=5x^{2} - 4\text{ and we can find that }\\&f(x+h)=5\cdot (h + x)^{2} - 4\\&f'(x)=\frac{5\cdot (h + x)^{2} - 4-(5x^{2} - 4)}{h}\\&f'(x)=\frac{10\cdot hx + 5\cdot h^{2}}{h}\\&f'(x)=5\cdot h + 10x\\&\text{Then, if we take the limit as h goes to zero, we find at }x=-9:\\&-90\end{align*}\)

The derivative of a function, \(\displaystyle f(x)\), as defined by the limit of the difference quotient is

\(\displaystyle \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\)

Taking the limit of our function - and remembering the limit at each step! - we get

\(\displaystyle \lim_{h\rightarrow 0}\frac{12(x+h)+3-(12x+3)}{h}=\lim_{h\rightarrow 0}\frac{12x+12h+3-12x-3}{h}=\lim_{h\rightarrow 0}\frac{12h}{h}=12\)

The derivative of a function, \(\displaystyle f(x)\), as defined by the limit of the difference quotient is

\(\displaystyle \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\)

Taking the limit of our function (and remembering to write the limit at each step) we get

\(\displaystyle \lim_{h\rightarrow 0}\frac{3(x+h)^2+(x+h)-(3x^2+x)}{h}=\lim_{h\rightarrow 0}\frac{3(x^2+2xh+h^2)+x+h-3x^2-x}{h}=\lim_{h\rightarrow 0}\frac{3x^2+6xh+3h^2+x+h-3x^2-x}{h}=\lim_{h\rightarrow 0}\frac{h(6x+3h+1)}{h}=6x+1\)

The derivative of a function \(\displaystyle f(x)\) is defined by the limit of the difference quotient, as follows:

\(\displaystyle \lim_{h\rightarrow 0}\frac{ f(x+h)-f(x)}{h}\)

Using this limit for our function, and remembering to write the limit at every step, we get

\(\displaystyle \lim_{h\rightarrow 0} \frac{5(x+h^2)+1-(5x^2+1)}{h}=\lim_{h\rightarrow 0} \frac{5(x^2+2xh+h^2)+1-5x^2-1}{h}=\lim_{h\rightarrow 0}\frac{10xh+5h^2}{h}=\lim_{h\rightarrow 0} (10x+5h) = 10x\)