What is the derivative of k(1-w/d e^{-sx}sin(dx+t)) ?

The derivative of k(1-w/d e^{-sx}sin(dx+t)) is -(kw(e^{-sx}dcos(dx+t)-se^{-sx}sin(dx+t)))/d

What is the first derivative of k(1-w/d e^{-sx}sin(dx+t)) ?

The first derivative of k(1-w/d e^{-sx}sin(dx+t)) is -(kw(e^{-sx}dcos(dx+t)-se^{-sx}sin(dx+t)))/d

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\frac{d}{d x} (k (1 - \frac{w}{d} e^{- s x} sin (d x + t)))

- \frac{k w ( e ^{- s x} d cos ( d x + t ) - s e ^{- s x} sin ( d x + t ) )}{d}

Solution steps

\frac{d}{d x} (k (1 - \frac{w}{d} e^{- s x} sin (d x + t)))

Simplify

k (1 - \frac{w}{d} e^{- s x} sin (d x + t)) : k (1 - \frac{e ^{- s x} w sin ( d x + t )}{d})

= \frac{d}{d x} (k (1 - \frac{e ^{- s x} w sin ( d x + t )}{d}))

Treat

s, k, w, d, t

as constants

Take the constant out:

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

= k \frac{d}{d x} (1 - \frac{e ^{- s x} w sin ( d x + t )}{d})

Apply the Sum/Difference Rule:

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

= k (\frac{d}{d x} (1) - \frac{d}{d x} (\frac{e ^{- s x} w sin ( d x + t )}{d}))

\frac{d}{d x} (1) = 0

\frac{d}{d x} (\frac{e ^{- s x} w sin ( d x + t )}{d}) = \frac{w ( - s e ^{- s x} sin ( d x + t ) + d e ^{- s x} cos ( d x + t ) )}{d}

= k (0 - \frac{w ( - s e ^{- s x} sin ( d x + t ) + d e ^{- s x} cos ( d x + t ) )}{d})

Simplify

k (0 - \frac{w ( - s e ^{- s x} sin ( d x + t ) + d e ^{- s x} cos ( d x + t ) )}{d}) : - \frac{k w ( e ^{- s x} d cos ( d x + t ) - s e ^{- s x} sin ( d x + t ) )}{d}

= - \frac{k w ( e ^{- s x} d cos ( d x + t ) - s e ^{- s x} sin ( d x + t ) )}{d}

- x

\frac{9 - x ^{2}}{16 x}

\int π x d x

\frac{d}{d x} (i n^{2} (x + 3))

\int_{0}^{4} 2 x + 1 d x

What is the derivative of k(1-w/d e^{-sx}sin(dx+t)) ?
The derivative of k(1-w/d e^{-sx}sin(dx+t)) is -(kw(e^{-sx}dcos(dx+t)-se^{-sx}sin(dx+t)))/d
What is the first derivative of k(1-w/d e^{-sx}sin(dx+t)) ?
The first derivative of k(1-w/d e^{-sx}sin(dx+t)) is -(kw(e^{-sx}dcos(dx+t)-se^{-sx}sin(dx+t)))/d