What is the (x^2-2x-2sin(x-2))^' ?

The (x^2-2x-2sin(x-2))^' is 2x-2-2cos(x-2)

What is the first (x^2-2x-2sin(x-2))^' ?

The first (x^2-2x-2sin(x-2))^' is 2x-2-2cos(x-2)

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(x^{2} - 2 x - 2 sin (x - 2))^{^{'}}

2 x - 2 - 2 cos (x - 2)

Solution steps

(x^{2} - 2 x - 2 sin (x - 2))^{'}

Apply the Sum/Difference Rule:

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

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

(x^{2})^{'} = 2 x

(2 x)^{'} = 2

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

= 2 x - 2 - 2 cos (x - 2)

n = 1 \sum \infty \frac{n !}{n !}

\int x cos (n x π) d x

\frac{\partial}{\partial x} (x arctan (\frac{y}{z}))

e x t re m e f (r) = e^{r} + x^{e}

\frac{d}{d y} (e^{x + y})

What is the (x^2-2x-2sin(x-2))^' ?
The (x^2-2x-2sin(x-2))^' is 2x-2-2cos(x-2)
What is the first (x^2-2x-2sin(x-2))^' ?
The first (x^2-2x-2sin(x-2))^' is 2x-2-2cos(x-2)