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Popular >

d/(dt)(acos(wt)+bsin(wt))

Solution

\frac{d}{d t} (a cos (wt) + b sin (wt))

Solution

- a w sin (wt) + b cos (wt) w

Solution steps

\frac{d}{d t} (a cos (wt) + b sin (wt))

Treat

a, b, w

as a constant

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (a cos (wt)) + \frac{d}{d t} (b sin (wt))

\frac{d}{d t} (a cos (wt)) = - a w sin (wt)

\frac{d}{d t} (b sin (wt)) = b cos (wt) w

= - a w sin (wt) + b cos (wt) w

Popular Examples

integral of (e^{5x})(cos(-2x))

\int (e^{5 x}) (cos (- 2 x)) d x

(dy)/(dx)=y-x,y(0)= 2/3

\frac{d y}{d x} = y - x, y (0) = \frac{2}{3}

integral of (2x+1)/((x-1)(x+3))

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

derivative of (x^2-x+5/(4x-1))

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

(\partial)/(\partial y)(e^{-2y}cos(x))

\frac{\partial}{\partial y} (e^{- 2 y} cos (x))

Frequently Asked Questions (FAQ)

What is the d/(dt)(acos(wt)+bsin(wt)) ?
The d/(dt)(acos(wt)+bsin(wt)) is -awsin(wt)+bcos(wt)w
What is the first d/(dt)(acos(wt)+bsin(wt)) ?
The first d/(dt)(acos(wt)+bsin(wt)) is -awsin(wt)+bcos(wt)w