What is the d/(dt)(3cos(t)-2e^{2t}) ?

The d/(dt)(3cos(t)-2e^{2t}) is -3sin(t)-4e^{2t}

What is the first d/(dt)(3cos(t)-2e^{2t}) ?

The first d/(dt)(3cos(t)-2e^{2t}) is -3sin(t)-4e^{2t}

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\frac{d}{d t} (3 cos (t) - 2 e^{2 t})

- 3 sin (t) - 4 e^{2 t}

Solution steps

\frac{d}{d t} (3 cos (t) - 2 e^{2 t})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (3 cos (t)) - \frac{d}{d t} (2 e^{2 t})

\frac{d}{d t} (3 cos (t)) = - 3 sin (t)

\frac{d}{d t} (2 e^{2 t}) = 4 e^{2 t}

= - 3 sin (t) - 4 e^{2 t}

(- 1, 0), (0, - 2)

(x^{2} + 3 x y + y^{2}) d x - x^{2} d y = 0

cos (x), - \frac{π}{2}

\frac{5}{x}, y = 3 x, y = \frac{1}{3} x

\int_{a}^{b} \frac{1}{x ^{2}} d x

What is the d/(dt)(3cos(t)-2e^{2t}) ?
The d/(dt)(3cos(t)-2e^{2t}) is -3sin(t)-4e^{2t}
What is the first d/(dt)(3cos(t)-2e^{2t}) ?
The first d/(dt)(3cos(t)-2e^{2t}) is -3sin(t)-4e^{2t}