Question

Quotient rule of sin3x.se X/3

Answer 1

`y' = (sec(x))/(3)[sin(3x) tan(x) + 3cos(3x)]`

Step-by-step explanation:

Given

`(sin3x\ secx)/(3)`

Required

The quotient rule

Quotient rule states that:

`y' = (V(du)/(dx) - U(dv)/(dx))/(v^2)`

Where

`v = 3` and
`(dv)/(dx) = 0`

`u = sin3x\ secx`

`(du)/(dx)` using product rule is:

`(du)/(dx) = sin3x * (d[secx])/(dx) + secx * (d[sin3x])/(dx)`

Differentiate all:

`(du)/(dx) = sin(3x) sec(x) tan(x) + 3sec(x)cos(3x)`

Factorize:

`(du)/(dx) = sec(x)[sin(3x) tan(x) + 3cos(3x)]`

So, the rule:

`y' = (V(du)/(dx) - U(dv)/(dx))/(v^2)` becomes

`y' = (3[sec(x)[sin(3x) tan(x) + 3cos(3x)]] - [sin3x\ secx] * 0)/(3^2)`

Solving further:

`y' = (3[sec(x)[sin(3x) tan(x) + 3cos(3x)]])/(9)`

`y' = (sec(x)[sin(3x) tan(x) + 3cos(3x)])/(3)`

`y' = (sec(x))/(3)[sin(3x) tan(x) + 3cos(3x)]`