Question

y=tan(5πt/2+π/6), then find dy/dt at t=0/meritnation

Answer 1

y = tan( 5πt/2 + π/6)

concept :- when, any function is in the form of
y = tan( ax + b )
then, differentiation is
dy/dx = sec²(ax + b) × a
use this here,

y = tan(5πt/2 + π/6)
dy/dt = sec²(5πt/2 + π/6) × 5π/2

now , put t = 0

dy/dt = 5π/2 × sec²( 5π×0/2 + π/6)

dy/dt = 5π/2 × sec²(π/6)

dy/dt = 5π/2 × {2/√3}²

dy/dt = 5π/2 × 4/3 = 10π/3

Answer 2

You simply have to apply chain rule!

y = tan(5πt/2+π/6)

dy/dt = d/dt( tan(5πt/2+π/6) )

dy/dt =

d/(d(5πt/2+π/6)) {tan(5πt/2+π/6)}×d(5πt/2+π/6)/dt

dy/dt = sec²(5πt/2+π/6)×(5π/2)

now the derivative at t=0 can be easily found out by putting t=0 in the expression lf dy/dt

dy/dt = sec²(π/6)× 5π/2 =(4/3)×(5π/2)
= 10π/3