What is the d/(dt)((w)(5cos^2(t)-10sin^2(t))) ?

The d/(dt)((w)(5cos^2(t)-10sin^2(t))) is -30wcos(t)sin(t)

What is the first d/(dt)((w)(5cos^2(t)-10sin^2(t))) ?

The first d/(dt)((w)(5cos^2(t)-10sin^2(t))) is -30wcos(t)sin(t)

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\frac{d}{d t} ((w) (5 cos^{2} (t) - 10 sin^{2} (t)))

- 30 w cos (t) sin (t)

Solution steps

\frac{d}{d t} ((w) (5 cos^{2} (t) - 10 sin^{2} (t)))

Treat

w

as a constant

Take the constant out:

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

= w \frac{d}{d t} (5 cos^{2} (t) - 10 sin^{2} (t))

Apply the Sum/Difference Rule:

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

= w (\frac{d}{d t} (5 cos^{2} (t)) - \frac{d}{d t} (10 sin^{2} (t)))

\frac{d}{d t} (5 cos^{2} (t)) = - 10 cos (t) sin (t)

\frac{d}{d t} (10 sin^{2} (t)) = 20 sin (t) cos (t)

= w (- 10 cos (t) sin (t) - 20 sin (t) cos (t))

Simplify

= - 30 w cos (t) sin (t)

(2 - 10 ln (\frac{4}{3}))^{2}

25 \cdot 5^{2 k}

\frac{d ^{2}}{d x ^{2}} (4 x (x^{2} + 9) \cdot (x - 9)) + (x^{2} + 9)^{2}

\frac{( x ^{2} - 5 ) ( x - 4 )}{( x + 1 ) ( 2 x ^{2} - 5 )}

(sin (x) + cos (x)) e^{x}

What is the d/(dt)((w)(5cos^2(t)-10sin^2(t))) ?
The d/(dt)((w)(5cos^2(t)-10sin^2(t))) is -30wcos(t)sin(t)
What is the first d/(dt)((w)(5cos^2(t)-10sin^2(t))) ?
The first d/(dt)((w)(5cos^2(t)-10sin^2(t))) is -30wcos(t)sin(t)