What is the derivative of sin(2θ)cos(θ) ?

The derivative of sin(2θ)cos(θ) is 2cos(2θ)cos(θ)-sin(θ)sin(2θ)

What is the first derivative of sin(2θ)cos(θ) ?

The first derivative of sin(2θ)cos(θ) is 2cos(2θ)cos(θ)-sin(θ)sin(2θ)

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derivative of

sin (2 θ) cos (θ)

2 cos (2 θ) cos (θ) - sin (θ) sin (2 θ)

Solution steps

\frac{d}{d θ} (sin (2 θ) cos (θ))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = sin (2 θ), g = cos (θ)

= \frac{d}{d θ} (sin (2 θ)) cos (θ) + \frac{d}{d θ} (cos (θ)) sin (2 θ)

\frac{d}{d θ} (sin (2 θ)) = cos (2 θ) \cdot 2

\frac{d}{d θ} (cos (θ)) = - sin (θ)

= cos (2 θ) \cdot 2 cos (θ) + (- sin (θ)) sin (2 θ)

Simplify

= 2 cos (2 θ) cos (θ) - sin (θ) sin (2 θ)

x \to - 2 + lim (3 x + 6 - 6)

\int x^{2} (2 x^{3} - 1)^{4} d x

\int \frac{1}{x} + (x^{5} e^{x}) d x

4 sin (5 t)

\frac{\partial}{\partial x} (x^{4})

What is the derivative of sin(2θ)cos(θ) ?
The derivative of sin(2θ)cos(θ) is 2cos(2θ)cos(θ)-sin(θ)sin(2θ)
What is the first derivative of sin(2θ)cos(θ) ?
The first derivative of sin(2θ)cos(θ) is 2cos(2θ)cos(θ)-sin(θ)sin(2θ)