What is the derivative of 152sin^2(x)?
回答 (12)
2020-10-06 7:37 pm
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
:)>
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
:)>
2020-10-06 8:56 pm
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).
Then g'(x) = 2sin(x)cos(x) = sin(2x);
Now f(x) = 152g(x);
Therefore f'(x) = 152g'(x) = 152sin(2x).
2020-10-06 7:01 pm
y = 152SinXSinX
Apply the Product Rule
dy/dx = 152 (SinXCosX + CosXSin)
dy/dx = 152(2SinXCosX)
dy/dx = 304SinXCosX
Done!!!!
Apply the Product Rule
dy/dx = 152 (SinXCosX + CosXSin)
dy/dx = 152(2SinXCosX)
dy/dx = 304SinXCosX
Done!!!!
2020-10-06 6:23 pm
D(152sin^2(x)) = 152sin(2x) .
2020-10-07 10:36 am
y = 152sin^2(x)
dy/dx = 304sin(x) d/dx cos(x)
= 304sin(x)cos(x) Answer//
dy/dx = 304sin(x) d/dx cos(x)
= 304sin(x)cos(x) Answer//
2020-10-07 7:44 am
d/dx(152 sin^2(x)) = 304 sin(x) cos(x)
2020-10-06 7:50 pm
d/dx(152sin^2x) = 304sinx cosx
2020-10-08 11:58 am
USE THE PRODUCT RULE FORMULA
2020-10-08 4:26 am
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)
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)
2020-10-07 10:06 pm
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))
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))
2020-10-06 6:32 pm
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
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
收錄日期: 2021-04-24 08:02:11
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