What is the derivative of-2sin(xcos(x)at[0.2pi]) ?

The derivative of-2sin(xcos(x)at[0.2pi]) is -1.25663…atcos(2x)

What is the first derivative of-2sin(xcos(x)at[0.2pi]) ?

The first derivative of-2sin(xcos(x)at[0.2pi]) is -1.25663…atcos(2x)

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\frac{d}{d x} (- 2 sin (x) cos (x) a t [0.2 π])

- 1.25663 \dots a t cos (2 x)

Solution steps

\frac{d}{d x} (- 2 sin (x) cos (x) a t (0.2 π))

Simplify

- 2 sin (x) cos (x) a t (0.2 π) : - 1.25663 \dots a t sin (x) cos (x)

= \frac{d}{d x} (- 1.25663 \dots a t sin (x) cos (x))

Treat

a, t

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= - 1.25663 \dots a t \frac{d}{d x} (sin (x) cos (x))

Apply the Product Rule:

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

f = sin (x), g = cos (x)

= - 1.25663 \dots a t (\frac{d}{d x} (sin (x)) cos (x) + \frac{d}{d x} (cos (x)) sin (x))

\frac{d}{d x} (sin (x)) = cos (x)

\frac{d}{d x} (cos (x)) = - sin (x)

= - 1.25663 \dots a t (cos (x) cos (x) + (- sin (x)) sin (x))

Simplify

- 1.25663 \dots a t (cos (x) cos (x) + (- sin (x)) sin (x)) : - 1.25663 \dots a t cos (2 x)

= - 1.25663 \dots a t cos (2 x)

\frac{18}{( s + 1 ) ^{2} ( s ^{2} + 4 )}

\frac{\partial}{\partial z} (y - z)

x \to 5 lim (\frac{1}{x ^{2} + 1})

g (x) = x^{2} + 6 x^{3}

\frac{\partial}{\partial x} (7 x^{2} y + 6 x y^{3})

What is the derivative of-2sin(xcos(x)at[0.2pi]) ?
The derivative of-2sin(xcos(x)at[0.2pi]) is -1.25663…atcos(2x)
What is the first derivative of-2sin(xcos(x)at[0.2pi]) ?
The first derivative of-2sin(xcos(x)at[0.2pi]) is -1.25663…atcos(2x)