To use this in the formula [latex]f'(x) = \frac{f(x+h) - f(x)}{h}[/latex], first we need to replace the [latex]f(x+h)[/latex] part of the formula. This is the same as [latex]f(x)[/latex] which is [latex]3x+5[/latex], except we replace [latex]x[/latex] with that [latex](x+h)[/latex] in parantheses. Like the following. The colors are only to highlight the substitution of [latex]f(x+h)[/latex] and [latex]f(x)[/latex]. We’ll drop the colors as soon as we need to combine expressions.
\begin{align*}
f'(x) & = \lim_{h \to 0} \frac{{\color{blue} f(x + h)} – {\color{red} f(x)}}{h} \\
& = \lim_{h \to 0} \frac{{\color{blue} 3(x + h) + 5} – {\color{red} 3x + 5}}{h} \\
\end{align*}
Now we continue to simplify and find the answer.
\begin{align*}
f'(x) & = \lim_{h \to 0} \frac{{\color{blue} f(x + h)} – {\color{red} f(x)}}{h} \\
& = \lim_{h \to 0} \frac{{\color{blue} 3(x + h) + 5} – {\color{red} 3x + 5}}{h} \\
& = \lim_{h \to 0} \frac{3x + 3h + 5 – 3x – 5}{h} \\
& = \lim_{h \to 0} \frac{3h}{h} \\
& = \lim_{h \to 0} 3 \\
& = \boxed{3}
\end{align*}
Here, we have [latex]f'(x) = 3[/latex]. That makes sense if you think about it: [latex]3x + 5[/latex] is a line with slope [latex]3[/latex]!

\begin{align*}
f'(x) & = \lim_{h \to 0} \frac{{\color{blue} f(x + h)} – {\color{red} f(x)}}{h} \\
& = \lim_{h \to 0} \frac{{\color{blue} (x + h)^2} – {\color{red} x^2}}{h} \\
& = \lim_{h \to 0} \frac{{\color{blue} x^2 + 2xh + h^2} – {\color{red} x^2}}{h} \\
& = \lim_{h \to 0} \frac{2xh + h^2}{h} \\
& = \lim_{h \to 0} \frac{h(2x + h)}{h} \\
& = \lim_{h \to 0} 2x + h \\
& = 2x + (0) \\
& = \boxed{2x}
\end{align*}

So what does this mean? Well, this means we double [latex]x[/latex] to find the slope of the tangent line of [latex]f(x) =x^2[/latex]. So at [latex]x = 3[/latex], the slope is [latex]6[/latex], and at [latex]x = 1.2[/latex], the slope is [latex]2.4[/latex]. ETC.