What is the d/(dt)(bsin(wt)) ?

The d/(dt)(bsin(wt)) is bcos(wt)w

What is the first d/(dt)(bsin(wt)) ?

The first d/(dt)(bsin(wt)) is bcos(wt)w

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\frac{d}{d t} (b sin (wt))

b cos (wt) w

Solution steps

\frac{d}{d t} (b sin (wt))

Treat

b, w

as a constant

Take the constant out:

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

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

Apply the chain rule:

cos (wt) \frac{d}{d t} (wt)

= cos (wt) \frac{d}{d t} (wt)

\frac{d}{d t} (wt) = w

= b cos (wt) w

\frac{d}{d x} (w \cdot sin (w c t) \cdot cos (w x))

\frac{\partial}{\partial x} (\frac{3}{x})

\frac{d x}{d t} = - 3 x (x^{2} - 1)

\int \frac{sin ( \frac{π}{x} )}{x ^{2}} d x

\frac{d y}{d t} = \frac{t ^{2}}{y ( 2 + t ^{3} )}