Differentiate

A) y = -Cos2x

B) y = Sin2tetta - 2Cos2tetta

C) f(tetta) = negative pie*Sin(2tetta - pie)

Tetta means that zero with a line going horizontally across mid zero. Like In trigonometric identities you have ___ tetta....

Go on: wolframalpha dot com

When page be open in rectangle type:

derivative cos(2x)

and click option =

After few seconds when you see result click option:

Show steps

Then in in rectangle type:

derivative sin(2theta)-2cos(2theta)

and click option =

then Show steps

C)

sin ( theta - pi ) = - sin ( theta )

sin ( 2 theta - pi ) = - sin ( 2 theta )

f(theta) = - pi * sin( 2theta - pi ) =

- pi * [ -sin ( 2 theta ) ]=

pi * sin ( 2 theta )

A) y'= 2sin(2x)

B) y'= 4sin(2theta) -2cos(2theta)
C) y'= 2piCos(2theta)

A) y'= - 2sin(2x)

B) y'= 4 sin ( 2 theta ) + 2 cos ( 2theta ) =

2 [ 2 sin (2 theta ) + cos ( theta ) ]

C) y'= 2piCos(2theta)

To differentiate each of the given equations, we can use the chain rule, product rule, and trigonometric differentiation rules. Let's differentiate each equation step by step:

A) y = -cos(2x)

To differentiate y with respect to x, we can use the chain rule:
dy/dx = d(-cos(2x))/dx

Applying the chain rule, we differentiate the outer function (-cos(2x)) first and then multiply it by the derivative of the inner function (2x):
dy/dx = -(-sin(2x) * 2)
= 2sin(2x)

Therefore, the derivative of y = -cos(2x) is dy/dx = 2sin(2x).

B) y = sin(2θ) - 2cos(2θ)

Similar to the previous equation, we will differentiate each term separately using the differentiation rules:
dy/dθ = d(sin(2θ))/dθ - d(2cos(2θ))/dθ

Using the chain rule, the derivative of sin(2θ) is cos(2θ) multiplied by the derivative of the inner function (2θ):
dy/dθ = cos(2θ) * 2 - (-2sin(2θ) * 2)

Simplifying, we get:
dy/dθ = 2cos(2θ) + 4sin(2θ)

Therefore, the derivative of y = sin(2θ) - 2cos(2θ) is dy/dθ = 2cos(2θ) + 4sin(2θ).

C) f(θ) = -πsin(2θ - π)

To differentiate f with respect to θ, we'll use the chain rule and the trigonometric differentiation rules:
df/dθ = d(-πsin(2θ - π))/dθ

Applying the chain rule, the derivative of -πsin(2θ - π) is (-πcos(2θ - π)) multiplied by the derivative of the inner function (2θ - π):
df/dθ = -πcos(2θ - π) * 2

Simplifying, we get:
df/dθ = -2πcos(2θ - π)

Therefore, the derivative of f(θ) = -πsin(2θ - π) is df/dθ = -2πcos(2θ - π).

In summary:
A) dy/dx = 2sin(2x)
B) dy/dθ = 2cos(2θ) + 4sin(2θ)
C) df/dθ = -2πcos(2θ - π)