Answer:
[tex]y' = \frac{sec(x)}{3}[sin(3x) tan(x) + 3cos(3x)][/tex]
Explanation:
Given
[tex]\frac{sin3x\ secx}{3}[/tex]
Required
The quotient rule
Quotient rule states that:
[tex]y' = \frac{V\frac{du}{dx} - U\frac{dv}{dx}}{v^2}[/tex]
Where
[tex]v = 3[/tex] and [tex]\frac{dv}{dx} = 0[/tex]
[tex]u = sin3x\ secx[/tex]
[tex]\frac{du}{dx}[/tex] using product rule is:
[tex]\frac{du}{dx} = sin3x * \frac{d[secx]}{dx} + secx * \frac{d[sin3x]}{dx}[/tex]
Differentiate all:
[tex]\frac{du}{dx} = sin(3x) sec(x) tan(x) + 3sec(x)cos(3x)[/tex]
Factorize:
[tex]\frac{du}{dx} = sec(x)[sin(3x) tan(x) + 3cos(3x)][/tex]
So, the rule:
[tex]y' = \frac{V\frac{du}{dx} - U\frac{dv}{dx}}{v^2}[/tex] becomes
[tex]y' = \frac{3[sec(x)[sin(3x) tan(x) + 3cos(3x)]] - [sin3x\ secx] * 0}{3^2}[/tex]
Solving further:
[tex]y' = \frac{3[sec(x)[sin(3x) tan(x) + 3cos(3x)]]}{9}[/tex]
[tex]y' = \frac{sec(x)[sin(3x) tan(x) + 3cos(3x)]}{3}[/tex]
[tex]y' = \frac{sec(x)}{3}[sin(3x) tan(x) + 3cos(3x)][/tex]
