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What is the derivative of 2lnx?
The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/x
What is the derivative of 3cosx?
The derivative of 3cos(x) is -3sin(x). This can be found using the chain rule, which states that the derivative of a composition of functions is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. In this case, the derivative of cos(x) is -sin(x), and when multiplied by the constant 3, we get -3sin(x) as the derivative of 3cos(x).
What is the derivative of csc x?
The derivative of csc(x) is -cot(x)csc(x).
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
What is the derivative of sec squared x?
derivative of sec2(x)=2tan(x)sec2(x)
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 derivative of sin under root theta?
The derivative of (sin (theta))^.5 is (cos(theta))/(2sin(theta))
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)
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) ]
Prove identity sin4x equals 4sinxcosx times 1 minus 2sin squared x?
sin4x=(4sinxcosx)(1-2sin^2x)
What is the period of the function y equals 2sin x?
2 pi
Integral of sin 2x divided by 1 plus cosine squared x?
3
1-2sin squared x equals?
3
What is the derivative of 2lnx?
The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/x
Derivative of cosx?
The derivative of cos(x) is negative sin(x). Also, the derivative of sin(x) is cos(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 range of the function y 2sin x?
The answer will range between '2' & '-2' Reason; The Sine function ranges between '1' & '-1' , so if it has a coefficient of '2', this will increase the range to '2' & '-2'.
