how do i find the derivative of f(x)= (x/2)-(sin2x/4)?

Answer 1

derivative of x is 1
derivative of sin2x = derivative of sin2x * derivative of 2x
= cos2x*2
combining the above
f'x = 1/2 - cos2x*2/4
= 1/2 - cos2x/2
= 1/2(1-cos2x)

Answer 2

If you meant f(x)=(x/2) - sin(2x/4) then diff the parts separately to get:
f'(x)=(1/2) - (1/2)cos(2x/4)

If you meant f(x)=(x/2) - (sin2x)/4 then also diff each part to get:
f'(x) = (1/2) - (1/2)cos2x
The (x/2) part is easy to diff, and if you're having problems with the trig part, use the chain rule.

Answer 3

For the first term, use the power rule: d/dx[x^n] = nx^(n-1), with n=1.

For the second term, use the chain rule d/dx[f(g(x))] = f'(g(x))g'(x), with f(x) = sin(x) and g(x)=2x.

You get f'(x) = 1/2 - 2cos2x/4 = (1-cos2x)/2

Answer 4

f(x) = x/2 - sin 2x/4

Simplify 2x/4 = x/2
f(x) = x/2 - sin x/2

Get the derivative of each term
f'(x) = 1/2 - cos x/2 (1/2)

Or simply
f'(x) = 1/2 - 1/2 cos x/2

^_^

Answer 5

The brackets imply that this is x/2-(sin(2x)/4) but you are maybe talking about x/2-sinxsinx/4 or even x/2 - sin(x/4)sin(x/4) or x/2-sin(x/2)

I'll do the easiest : f(x)= x/2-sin(x/2)
substitute y=x/2. dy/dx=1/2.
f'(y)=1-cos(y) , f'(x)=(1-cos(x/2))/2

Answer 6

Damn easy!
f'(x)= 1/2- 1/4(cos 2x).2
= 1/2- 1/2(cos2x)
= 1/2(1-cos2x)

Answer 7

The answer is (1/2)-(cos(2x))/2
Just work it backwards using your Basic differentiation rules. d/dx [sin u]=(cos u)u' and d/dx[u/v] = (vu'-uv')/v^2 So Fill the numbers in for those two rules.

Answer 8

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Answer 9

f(x)'=1/2-1/2cos2x