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What is the second derivative of y equals 2sinx cosx?
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)
What is the average rate of change from x equals 2 to x equals 3 for the function f of x equals 3 ln x?
It is 1.2164
Integral of sin 2x divided by 1 plus cosine squared x?
3
How do you know that y is a function of x?
If you plug in y for the x function, and it equals the answer you got, it is right.
Is x plus y equals 4 a function?
Yes.
What is the range of the function y 2sin x?
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.
Prove identity sin4x equals 4sinxcosx times 1 minus 2sin squared x?
sin4x=(4sinxcosx)(1-2sin^2x)
1-2sin squared x equals?
3
Solve 3sin x equals 2 for 0x360?
3 sin(x) = 2sin(x) = 2/3x = 41.81 degreesx = 138.19 degrees
Sin2x plus cosx equals 0?
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.
If tan y equals to x where y is acute find 2 sin y in terms of x?
2sin(y) = 2x/sqrt(1+x^2)
Why is 2 the answer for amplitude of y equals -2 sin x?
4
What is the second derivative of y equals 2sinx cosx?
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)
Express sin 2x as a function of sin x?
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)
2 sin x plus 1 equals 3 cos x - 1?
2sin(x+1)=3cos(x-1)Since there is only one variable in this equation, it cannot be implicitly differentiated.
What is the period of y equals tan2x?
The period of the tangent function is PI. The period of y= tan(2x) is PI over the coefficient of x = PI/2
How do you solve the following equation using identities sin2x-sinx equals 0?
2
