Q: What is the derivative of 5 sin2x?

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sin2X = sin2X What is it about ' equation ' you do you not understand. Of course they are equal!

(1 - tan2x)/(1 + tan2x) = (1 - sin2x/cos2x)/(1 + sin2x/cos2x) = (cos2x - sin2x)/(cos2x + sin2x) = (cos2x - sin2x)/1 = (cos2x - sin2x) = cos(2x)

sin2x because sin2x + cos2x = 1

The "double prime", or second derivative of y = 5x, equals zero. The first derivative is 5, a constant. Since the derivative of any constant is zero, the derivative of 5 is zero.

Use the formula for the derivative of a power. The square root of (x-5) is the same as (x-5)1/2.

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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.

f'(x)=-sin2x(2) f'(x)=-2sin2x First do the derivative of cos u, which is -sin u. Then because of the chain rule, you have to take the derivative of what's inside and the derivative of 2x is 2.

y=1/sinxy'=(sinx*d/dx(1)-1*d/dx(sinx))/(sin2x)y'=(sinx*0-1(cosx))/(sin2x)y'=(-cosx)/(sin2x)y'=-(cosx/sinx)*(1/sinx)y'=-cotx*cscx

sin2X = sin2X What is it about ' equation ' you do you not understand. Of course they are equal!

-4

(1 - tan2x)/(1 + tan2x) = (1 - sin2x/cos2x)/(1 + sin2x/cos2x) = (cos2x - sin2x)/(cos2x + sin2x) = (cos2x - sin2x)/1 = (cos2x - sin2x) = cos(2x)

Sin2x = radical 2

The derivative of 5x is 5.

You can take out any constant from a derivative. In other words, this is the same as 5 times the derivative of sec x.

sin2x because sin2x + cos2x = 1

1

The proof of this trig identity relies on the pythagorean trig identity, the most famous trig identity of all time: sin2x + cos2x = 1, or 1 - cos2x = sin2x. 1 + cot2x = csc2x 1 = csc2x - cot2x 1 = 1/sin2x - cos2x/sin2x 1 = (1 - cos2x)/sin2x ...using the pythagorean trig identity... 1 = sin2x/sin2x 1 = 1 So this is less of a proof and more of a verification.