what is the dy/dx of 3sin(squared) x?

Question

if the derivative of 3sinx is -3cosx, then what is the derivative of 3sin(squared) x?

Answer 1

Michaela R is right, [No, she isn't. She put a minus sign in, which is incorrect.] but would you like some explanation?
The derivative of 3 u^2 with respect to u is 3*2u, which is 6u.

If we want the derivative with respect to x, we must multiply this derivative (with respect to u) by du/dx.
In this case, du/dx = cos x when u = sin x [ No, you're wrong about that. The negative sign comes in when we differentiate cos, not sin].

Hence the derivative you need is
3*2sinx * cos x
= 6 sinx cos x, or, if you want to show off, 3 sin 2x

This can also be confirmed by using
3(sin x)^2 = (3/2)(1 - cos 2x), and the derivative of the right-hand side is
(3/2)(2 sin 2x)
= 3 sin 2x.

Answer 2

dy/dx 3 sin² x = 3(2)sin x cos x = 6 sin x cos x.
Here we have used the power rule and the chain rule.
But 2 sin x cos x = sin 2x, so this may also
be written as 3 sin 2x.

Answer 3

dy/dx of 3sin^2 (x)

y= 3(sin (x))^2
dy/dx= 2*3 (sinx)^1(der. of sinx)
dy/dx= 2*3 (sinx)^1(cosx)
dy/dx= 6sinxcosx
dy/dx=3(2sinxcosx)
dy/dx=3*sin2x

Answer 4

derivative of sin^2x=sin(2x)
cos^2(x)=-sin(2x)
remember these formulae.
the answer for question is 3sin(2x)
according to ur assumption the formulae and the answer given are additive inverse---- that is -3sin(2x)

Answer 5

thats not important

Answer 6

it's

-6sin(x)cos(x)