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

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

\int \frac{sec ^{2} ( x )}{1 + 2 tan ( x )} d x

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

\int (5 x^{4} + 64) arctan (\frac{x}{2}) d x

f (x) = 5 - x

lo g_{10} (\frac{( x + 1 ) ( x - 8 )}{( x - 1 ) ( x + 8 )})

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