What is the (sin(2pix)+2pixcos(2pix))^' ?

The (sin(2pix)+2pixcos(2pix))^' is 4picos(2pix)-4pi^2xsin(2pix)

What is the first (sin(2pix)+2pixcos(2pix))^' ?

The first (sin(2pix)+2pixcos(2pix))^' is 4picos(2pix)-4pi^2xsin(2pix)

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(sin (2 π x) + 2 π x cos (2 π x))^{^{'}}

4 π cos (2 π x) - 4 π^{2} x sin (2 π x)

Solution steps

(sin (2 π x) + 2 π x cos (2 π x))^{'}

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= (sin (2 π x))^{^{'}} + (2 π x cos (2 π x))^{^{'}}

(sin (2 π x))^{'} = cos (2 π x) \cdot 2 π

(2 π x cos (2 π x))^{'} = 2 π (cos (2 π x) - 2 π x sin (2 π x))

= cos (2 π x) \cdot 2 π + 2 π (cos (2 π x) - 2 π x sin (2 π x))

Simplify

cos (2 π x) \cdot 2 π + 2 π (cos (2 π x) - 2 π x sin (2 π x)) : 4 π cos (2 π x) - 4 π^{2} x sin (2 π x)

= 4 π cos (2 π x) - 4 π^{2} x sin (2 π x)

\frac{d y}{d x} = \frac{x - y}{x}

\frac{\partial}{\partial x} (6 x (x^{2} + 3 y^{2} - 1)^{2})

n = 1 \sum \infty cos (n)

x \to 0 lim (\frac{cos ( x )}{x})

\frac{\partial}{\partial x} (x^{2} - y^{2} + 5 x y)

What is the (sin(2pix)+2pixcos(2pix))^' ?
The (sin(2pix)+2pixcos(2pix))^' is 4picos(2pix)-4pi^2xsin(2pix)
What is the first (sin(2pix)+2pixcos(2pix))^' ?
The first (sin(2pix)+2pixcos(2pix))^' is 4picos(2pix)-4pi^2xsin(2pix)