2 pi
y = 2sin(x)cos(x)Use the product rule: uv' + vu' where u is 2sin(x) and v is cos(x) to find first derivative:y' = 2sin(x)(-sin(x)) + cos(x)2cos(x)Simplify:y' = 2cos2(x)-2sin2(x)y' = 2(cos2(x)-sin2(x))Use trig identity cos(2x) = cos2(x)-sin2(x):y' = 2cos(2x)Take second derivative using chain rule:y'' = 2(-sin(2x)cos(2x))Simplify:y'' = -2sin(2x)(2)Simplify:y'' = -4sin(2x)y'' = -4sin(2x)
It is 1.2164
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If you plug in y for the x function, and it equals the answer you got, it is right.
By Angle-Addition, cos(2x) = 2cos(x)^2-1 So, sin(x)cos(2x) = [2cos(x)^2-1]sin(x) = 2sin(x)cos(x)^2 - sin(x) Int[2sin(x)cos(x)^2 - sin(x)] = (-2/3)cos(x)^3 + cos(x) +K
y = 2sin(x)? If that's your function, well we know that sin(x) oscillates between y = 1 and y = -1, but in our case we have double that from 2sin(x), so our range is -2 to 2.
sin4x=(4sinxcosx)(1-2sin^2x)
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3 sin(x) = 2sin(x) = 2/3x = 41.81 degreesx = 138.19 degrees
sin 2x + cos x = 0 (substitute 2sin x cos x for sin 2x)2sin x cos x + cos x = 0 (divide by cos x each term to both sides)2sin x + 1 = 0 (subtract 1 to both sides)2sin x = -1 (divide by 2 to both sides)sin x = -1/2Because the period of the sine function is 360⁰, first find all solutions in [0, 360⁰].Because sin 30⁰ = 1/2 , the solutions of sin x = -1/2 in [0, 360] arex = 180⁰ + 30⁰ = 210⁰ (the sine is negative in the third quadrant)x = 360⁰ - 30⁰ = 330⁰ (the sine is negative in the fourth quadrant)Thus, the solutions of the equation are given byx = 210⁰ + 360⁰n and x = 330⁰ + 360⁰n, where n is any integer.
2sin(y) = 2x/sqrt(1+x^2)
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y = 2sin(x)cos(x)Use the product rule: uv' + vu' where u is 2sin(x) and v is cos(x) to find first derivative:y' = 2sin(x)(-sin(x)) + cos(x)2cos(x)Simplify:y' = 2cos2(x)-2sin2(x)y' = 2(cos2(x)-sin2(x))Use trig identity cos(2x) = cos2(x)-sin2(x):y' = 2cos(2x)Take second derivative using chain rule:y'' = 2(-sin(2x)cos(2x))Simplify:y'' = -2sin(2x)(2)Simplify:y'' = -4sin(2x)y'' = -4sin(2x)
The double angle formula states that sin(2x) = 2sin(x)cos(x), and because sin2(x) + cos2(x) = 1, sin(2x) = 2sin(x)*SQRT(1-sin2(x)). However, that will only give you the correct result when cos(x) is positive. (because SQRT only returns the positive square root)
The period of the tangent function is PI. The period of y= tan(2x) is PI over the coefficient of x = PI/2
2sin(x+1)=3cos(x-1)Since there is only one variable in this equation, it cannot be implicitly differentiated.
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