Why is the gradient of the curve of a strong base titrated with strong acid small up until equivalence?
Let's calculate the $p$H by changing the volume of $\ce{HCl}$ added. I will use the unit millimole ($1 mmol = 0.001 mol$), which is more practical than the mole.
In the very beginning the solution contains $50 mL~ \ce{NaOH}~ 0.1 M$, or $5 mmol$ $\ce{NaOH}$. Let's take, for example, the point when you have added $30  mL~ \ce{HCl} ~0.1 M$ (or $3.0 mmol$$ ˙\ce{HCl}$). After this addition, the solution contains $5 - 3.0 = 2.0 mmol$ NaOH in a volume $50 mL + 30 mL = 80 mL$. The concentration of $\ce{NaOH}$ is $\frac{2.0 mmol}{80 mL} = 0.025 M$. The concentration $\ce{[H+] = 10^{-14}/0.025 = 4 10^{-13}M}$. So $p$H = $12.4$. You see that the pH has fallen from $13.0$ to $12.4$ for more than half the trip to the equivalencce point. A change from $13.0$ to $12.4$ is rather small. The point of the curve is still nearly at its initial value. The gradient of pH between the beginning and here is small.
If you take the same calculation with an addition of $45 mL~ \ce{ NaOH}~ 0.1 M$, or $4.5 mmol$, the solution contains still $ 5 mmol - 4.5 mmol = 0.5 mmol ~\ce{NaOH}$ in $95 mL$. The concentration in $\ce{NaOH}$ is $\frac{0.5 mmol}{95 mL} = 5.26·10^{-3} M$. $\ce{[H^+] =  1,9·10^{-12}}$, and $p$H = $11.7$ We are near the equivalence point but the pH has not decreased a lot. It is still $11.7$. As you see the gradient (or the slope) of the curve is not important in the part of the curve between $0 mL$ and $45 mL$.
And suddenly between $45 mL$ and $50 mL$, the $p$H drops from $11.7$ to $7$. This is a huge change in just $5 mL$.
If you want you can calculate one or two other points along this titration. Try for example $49 mL$. You will be surprised ! It still gives a rather high $p$H value.