Question

whats the first derivative of y=sin (1+3x)^2

Answer 1

To solve this you need to use the "outside-inside" rule, or the chain rule.

Answer 2

did you get this one?
if this were just
y= sin(x)
dy = d sin(x)
dy= cos(x) dx
or
dy/dx = cos(x)

if we let u= (1+3x)^2
y= sin(u)
dy= cos(u) du
or (replace u)
dy= cos (1+3x)^2 d(1+3x)^2
can you finish?

Answer 3

the sine isn't being squared?

Answer 4

I am assuming the problem is
\[ \frac{d}{dx}\sin( (1+3x)^2) \]

if it is
\[ \frac{d}{dx}\sin^2( 1+3x) \]
that is a different problem