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What is the derivative of root sin2x?
You are supposed to use the chain rule for this. First step: derivative of root of sin2x is (1 / (2 root of sin 2x)) times the derivative of sin 2x. Second step: derivative of sin 2x is cos 2x times the derivative of 2x. Third step: derivative of 2x is 2. Finally, you need to multiply all the parts together.
Find the derivative of y x2 sin x 2xcos x - 2sin x?
y = (x^2)(sin x)(2x)(cos x) - 2sin xy' = [[(x^2)(sin x)][(2x)(cos x)]]' - (2sin x)'y' = [[(x^2)(sin x)]'[(2x)(cos x)] + [(2x)(cos x)]'[(x^2)(sin x)]]- (2sin x)'y' = [[(x^2)'(sin x) + (sin x)'(x^2)][(2x)(cos x)] + [(2x)'(cos x) + (cos x)'(2x)][(x^2)(sin x)] ] - 2(cos x)y' = [[(2x)(sin x )+ (cos x)(x^2)][(2x)(cos x)] + [2cos x - (sin x)(2x)][(x^2)(sin x)]] - 2(cos x)y' = (4x^2)(sin x cos x) + (2x^3)(cos x)^2 + (2x^2)(sin x cos x) - (2x^3)(sin x)^2 - 2cos xy' = (6x^2)(sin x cos x) + (2x^3)(cos x)^2 - (2x^3)(sin x)^2 - 2cos x (if you want, you can stop here, or you can continue)y' = (3x^2)(2sin x cos x) + (2x^3)[(cos x)^2 - (sin x)^2] - 2cos xy' = (3x^2)(sin 2x) + (2x^3)(cos 2x) - 2 cos xy' = (2x^3)(cos 2x) + (3x^2)(sin 2x) - 2 cos x
How do you prove tan x plus tan x sec 2x equals tan 2x?
tan x + (tan x)(sec 2x) = tan 2x work dependently on the left sidetan x + (tan x)(sec 2x); factor out tan x= tan x(1 + sec 2x); sec 2x = 1/cos 2x= tan x(1 + 1/cos 2x); LCD = cos 2x= tan x[cos 2x + 1)/cos 2x]; tan x = sin x/cos x and cos 2x = 1 - 2 sin2 x= (sin x/cos x)[(1 - 2sin2 x + 1)/cos 2x]= (sin x/cos x)[2(1 - sin2 x)/cos 2x]; 1 - sin2 x = cos2 x= (sin x/cos x)[2cos2 x)/cos 2x]; simplify cos x= (2sin x cos x)/cos 2x; 2 sinx cos x = sin 2x= sin 2x/cos 2x= tan 2x
What is the derivative of 3cosx2?
I'm assuming your question reads "What is the derivative of 3cos(x2)?" You must use the Chain Rule. The derivative of cos(x2) equals -sin(x2) times the derivative of the inside (x2), which is 2x. So... d/dx[3cos(x2)] = -6xsin(x2)
What is the 87th derivative of sin x?
Every fourth derivative, you get back to "sin x" - in other words, the 84th derivative of "sin x" is also "sin x". From there, you need to take the derivative 3 more times, getting:85th derivative: cos x86th derivative: -sin x87th derivative: -cos x
What is the derivative of sin2x2?
I'm not sure what you're asking. The derivative of sin(2x^2) is 4xcos(2x^x)dx.The derivative of (sin(2x^2)^2) is 8xsin(2x^2)cos(2x^2)dx.
What is the derivative of 5 sin2x?
d/dx [ 5 sin(2x) ] = 10 cos (2x)
What is the derivative of root sin2x?
You are supposed to use the chain rule for this. First step: derivative of root of sin2x is (1 / (2 root of sin 2x)) times the derivative of sin 2x. Second step: derivative of sin 2x is cos 2x times the derivative of 2x. Third step: derivative of 2x is 2. Finally, you need to multiply all the parts together.
Deriative of cosine squared x?
d/dx (cos x)^2 using the rule of chain, take derivative of the external, times derivative of the internal = 2 (cos x)(-sin x) =-2sinx cos x = - sin(2x)
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)
Find the derivative of y x2 sin x 2xcos x - 2sin x?
y = (x^2)(sin x)(2x)(cos x) - 2sin xy' = [[(x^2)(sin x)][(2x)(cos x)]]' - (2sin x)'y' = [[(x^2)(sin x)]'[(2x)(cos x)] + [(2x)(cos x)]'[(x^2)(sin x)]]- (2sin x)'y' = [[(x^2)'(sin x) + (sin x)'(x^2)][(2x)(cos x)] + [(2x)'(cos x) + (cos x)'(2x)][(x^2)(sin x)] ] - 2(cos x)y' = [[(2x)(sin x )+ (cos x)(x^2)][(2x)(cos x)] + [2cos x - (sin x)(2x)][(x^2)(sin x)]] - 2(cos x)y' = (4x^2)(sin x cos x) + (2x^3)(cos x)^2 + (2x^2)(sin x cos x) - (2x^3)(sin x)^2 - 2cos xy' = (6x^2)(sin x cos x) + (2x^3)(cos x)^2 - (2x^3)(sin x)^2 - 2cos x (if you want, you can stop here, or you can continue)y' = (3x^2)(2sin x cos x) + (2x^3)[(cos x)^2 - (sin x)^2] - 2cos xy' = (3x^2)(sin 2x) + (2x^3)(cos 2x) - 2 cos xy' = (2x^3)(cos 2x) + (3x^2)(sin 2x) - 2 cos x
How do you prove tan x plus tan x sec 2x equals tan 2x?
tan x + (tan x)(sec 2x) = tan 2x work dependently on the left sidetan x + (tan x)(sec 2x); factor out tan x= tan x(1 + sec 2x); sec 2x = 1/cos 2x= tan x(1 + 1/cos 2x); LCD = cos 2x= tan x[cos 2x + 1)/cos 2x]; tan x = sin x/cos x and cos 2x = 1 - 2 sin2 x= (sin x/cos x)[(1 - 2sin2 x + 1)/cos 2x]= (sin x/cos x)[2(1 - sin2 x)/cos 2x]; 1 - sin2 x = cos2 x= (sin x/cos x)[2cos2 x)/cos 2x]; simplify cos x= (2sin x cos x)/cos 2x; 2 sinx cos x = sin 2x= sin 2x/cos 2x= tan 2x
Derivative of cosx?
The derivative of cos(x) is negative sin(x). Also, the derivative of sin(x) is cos(x).
How do you derive sin2x?
If you are refering to the double-angle formula for sin(x), the best way is to use what is known as Euler's identity. Euler's identity is eix = cos(x) + i*sin(x) where x is any real angle in radians, e is Euler's constant 2.71828182845... and i is the imaginary number: SQRT(-1). Assuming that is true, then ei(2x) = cos(2x) + i*sin(2x) and that is the same as saying (eix)2= cos(2x) + i*sin(2x) and substituting from the original equation: (cos(x) + i*sin(x))2 = cos(2x) + i*sin(2x). By distribution, remembering that i2 = -1, we get cos2(x) + i*2*sin(x)*cos(x) - sin2(x) = cos(2x) + i*sin(2x). Now we can separate the equation into its real and imaginary parts. cos2(x) - sin2(x) = cos(2x) and i*2*sin(x)*cos(x) = i*sin(2x), and after i cancels, there's our good old double angle formula. If derive refers to derivative, then use the chain rule. d(sin(2x))/dx=2cos(2x)
What is the derivative of 3cosx2?
I'm assuming your question reads "What is the derivative of 3cos(x2)?" You must use the Chain Rule. The derivative of cos(x2) equals -sin(x2) times the derivative of the inside (x2), which is 2x. So... d/dx[3cos(x2)] = -6xsin(x2)
What is sec2x in relationship to cos?
Sec(2x) = 1/Cos(2x)
What is the 87th derivative of sin x?
Every fourth derivative, you get back to "sin x" - in other words, the 84th derivative of "sin x" is also "sin x". From there, you need to take the derivative 3 more times, getting:85th derivative: cos x86th derivative: -sin x87th derivative: -cos x
