Question

if v=5t cube+4t square+8t +1 find dv/dt at t=1
​

Answer 1

$GOOD \: MORNING \: \\ \\ v = 5t {}^{3} + 4t {}^{2} + 8t + 1 \\ find \: \: dv \div dt \: \: at \: \: \: t \: \: = 1 \\ \\ we \: know \: that \: \\ dx {}^{n} \div dx = nx {}^{n - 1} \\ \\ dv \div dt = 3 \times 5t {}^{3 - 1} + 2 \times 4t {}^{2 - 1} + 8 \\ \\ dv \div dt = 15t {}^{2} + 8t + 8 \\ \\ at \: \: t = 1 \\ \\ dv \div dt = 15 + 16 \\ \\ dv \div dt = 31ms {}^{ - 2}$