Solusi Soal Nomor 5 Uji Sertifikasi Pendidikan Menengah

$latex {\bf 5.\ \ (a)}\\\\{\largetext \begin{array}{rcl}\frac{d}{dx}\Big(\frac{3x+4}{x-2}\Big)&=&\frac{\frac{d}{dx}(3x+4).(x-2)-(3x+4).\frac{d}{dx}(x-2)}{(x-2)^2}\\\\&=&\frac{3.(x-2)-(3x+4).1}{(x-2)^2}\\\\&=&\frac{3x-6-3x-4}{(x-2)^2}\\\\&=&\frac{-10}{(x-2)^2}\\\\&=&-\frac{10}{(x-2)^2}\end{array}}$

$latex {\bf 5.\ \ (b)}\\y=3x^2+5x-12\quad \Rightarrow \quad \frac{dy}{dx}=6x+5\\\\Untuk\ x=2\ diperoleh\ gradien\ garis\ singgung\ kurva:\\\\\begin{matrix} & \ \end{matrix}m=\frac{dy}{dx}=6.2+5=12+5=17\\\\(i)\ \ \ Persamaan\ garis\ singgung\ kurva\ di\ titik\ P(2,10)\ adalah\\\begin{matrix} & \ \end{matrix}\begin{array}{rcl}y-10&=&17(x-2)\\y-10&=&17x-34\\y&=&17x-24\end{array}\\\\(ii)\ \ Persamaan\ garis\ normal\ kurva\ di\ titik\ P(2,10)\ adalah\\\\\begin{matrix} & \ \end{matrix}\begin{array}{rcl}y-10&=&-\frac{1}{17}(x-2)\\\\17y-170&=&-x+2\\x+17y-172&=&0\end{array}$ $latex {\bf 5.\ \ (c)}\\Luas\ daerah\ persegi:\quad L=x^2\quad \Rightarrow \quad \frac{dL}{dx}=2x\\\\\begin{array}{rcl}\frac{dL}{dt}&=&\frac{dL}{dx}\times \frac{dx}{dt}\\\\&=&2x\times 4\\&=&8x\end{array}\\\\Laju\ perubahan\ luas\ daerahnya\ ketika\ panjang\ sisinya\ 5\ cm\ adalah\\\\\begin{matrix} & \ \end{matrix}\begin{array}{rcl}\frac{dL}{dt}&=&8\times 5\\&=&40\end{array}$