What is the derivative of 152sin^2(x)?

The answer is as follows:

cos2x = 1 - 2sin²x

so, sin²x = 1/2 - (cos2x)/2

Then, d(sin²x)/dx = sin2x

Hence, d(152sin²x)/dx = 152sin2x

As sin2x = 2sinxcosx, we can say:

152(2sinxcosx) => 304sinxcosx

Note: the former is a bit neater

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Put g(x) = sin^2(x);
Then g'(x) = 2sin(x)cos(x) = sin(2x);
Now f(x) = 152g(x);
Therefore f'(x) = 152g'(x) = 152sin(2x).

y = 152SinXSinX 
Apply the Product Rule 
dy/dx = 152 (SinXCosX + CosXSin) 
dy/dx = 152(2SinXCosX)
dy/dx = 304SinXCosX
Done!!!!

D(152sin^2(x)) = 152sin(2x) .

y = 152sin^2(x)
dy/dx = 304sin(x) d/dx cos(x)
 = 304sin(x)cos(x) Answer//

d/dx(152 sin^2(x)) = 304 sin(x) cos(x)

d/dx(152sin^2x) = 304sinx cosx

USE THE PRODUCT RULE FORMULA

let u = sin(x)
then your expression becomes
y=152u^2

and hence dy/dx = dy/du * du/dx
dy/du = 152 * 2u = 304u (==> 304sin(x))
du/dx = cos(x)

so dy/dx = 304sin(x)cos(x)

Let f(x)=152sin^2(x), then
f '(x)=d[152sin^2(x)]dx
=>
f '(x)=152(2)sin(x)d[sin(x)/dx]
(the chain rule)
=>
f '(x)=304sin(x)cos(x)
=>
f '(x)=152sin(2x)
(sin(2x)=2sin(x)cos(x))

The result will be 152*d/dx{[sin(x)]^2}

Method 1: Treat [sin(x)]^2 as the product sin(x)* sin(x) 
d/dx{[sin(x)]^2} = sin(x)cos(x) + cos(x)sin(x) = 2sin(x)cos(x) ...(1)

Method 2: Use sin(x)]^2 = 1 - cos(2x)
d/dx{[sin(x)]^2} = 0 – d/dx[cos(2x)] = sin(2x) ..........................(2)

and of course (1) and (2) are equivalent. Multiply either by 152