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y = (x^2)(sin x)(2x)(cos x) - 2sin x
y' = [[(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 x
y' = (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 x
y' = (3x^2)(sin 2x) + (2x^3)(cos 2x) - 2 cos x
y' = (2x^3)(cos 2x) + (3x^2)(sin 2x) - 2 cos x
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What is the derivative of -2sin x?
The derivative of sin(x) is cos(x). Coefficients act like constants and always remain in derivatives. So, the derivative is -2cos(x).
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 sin?
Generally, the derivative of sine is cosine.
How do you find the derivative of y equals sin8x?
Derivative of sin x = cos x, so chain rule to derive 8x = 8 , answer is 8cos8x
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.
What is the derivative of sin under root theta?
The derivative of (sin (theta))^.5 is (cos(theta))/(2sin(theta))
What is the derivative of -2sin x?
The derivative of sin(x) is cos(x). Coefficients act like constants and always remain in derivatives. So, the derivative is -2cos(x).
Derivative of 1 plus cos2x?
First find the derivative of each term. The derivative of any constant is zero, so d(1)/dx=0. To find the derivative of cos2x, use the chain rule. d(cos2x)/dx=-sin(2x)(2)=-2sin(2x) So the answer is 0-2sinx, or simply -2sinx
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)
How do you differentiate 9 to the power sin squared x sorry I wasn't allowed to use mathematical notation when asking?
derivative of 9[sin(x)]^2 is found by first letting u(x)=[sin(x)]^2. Note that sin2x = [sin(x)]^2, and the ^2 means raising the base to the exponent 2. Find the d(9u(x))/dx using the chain rule. d( 9u(x) )/dx = (d(9u)/du)(du/dx ) , by the chain rule. So we need: d(9u)/du = 9ulog(9) du/dx = d( [sin(x)]^2 )/dx = 2sin(x) d( sin(x) )/dx = 2sin(x)cos(x) Puttin this together gives: d( 9u(x) )/dx = 9u(log(9)) 2sin(x)cos(x) Now substitute in u(x) = [sin(x)]^2. d( 9u(x) )/dx = 9[sin(x)]^2(log(9)) 2sin(x)cos(x) = 2 log(9) 9[sin(x)]^2sin(x)cos(x) or = log(9) 9[sin(x)]^2sin(2x)
How do you finish the equation sin 2 theta?
(/) = theta sin 2(/) = 2sin(/)cos(/)
Find the second derivative of y equals sin sqaured x?
y = sin2(x) y' = 2sin(x)cos(x) y'' = 2 [ cos(x)cos(x) + sin(x)(-sin(x)) ] = 2 [ cos2(x) - sin2(x) ] = 2 [ 1 - sin2(x) - sin2(x) ] = 2 [ 1 - 2sin2(x) ]
Find derivative of sinx using first principle?
The derivative of sin(x) is cos(x).
What does 2sin squared theta equal and why?
It is 2*sin(theta)*sin(theta) because that is how multiplication is defined!
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
Why is 2 sin theta cos theta equal to sin 2theta?
because sin(2x) = 2sin(x)cos(x)
How do you solve 2 sin squared x is equal to sin x?
There will be 4 possible solutions if all you are looking for is the angles, so you will need to find out which quadrant your angle is in.2sin²x = sin x **Subtract sin x from both sides.2sin²x - sin x = 0 **Then factor out sin x.sin x(2sin x - 1) = 0 **Set each equal to zero. (AB=0 is the same as A=0 OR B=0).sin x = 0 or 2sin x - 1 = 0 to 2sin x = 1 to sin x = 1/2At this point all that is left to do is find out where sin x = 0 or 1/2, which is 0, 180 for sin x = 0 or 30, 150 for sin x = 1/2.
