ข้อสอบคณิต 9 สามัญ - ปี 2563

$$\begin{gathered} \mathrm{จากโจทย์} \\ f\left(\sqrt{x}-1\right) = x \\ \mathrm{ให้} \sqrt{x}-1 = k \\ \sqrt{x} = k+1 \\ x = {\left(k+1\right)}^2 \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} f\left(k\right) = {\left(k+1\right)}^2 \\ f\left(k\right) = k^2+2k+1 \\ f\text{'}\left(k\right) = 2k+2 \\ f\text{'}\left(1\right) = 2\left(1\right)+2 = 4 \\ \mathrm{ดังนั้น} \mathrm{ตอบ} 4 \end{gathered}$$