What is the derivative of e^{,\at}sin(bt) ?

The derivative of e^{,\at}sin(bt) is e^,cos(bt)t

What is the first derivative of e^{,\at}sin(bt) ?

The first derivative of e^{,\at}sin(bt) is e^,cos(bt)t

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derivative of

e^{, at} sin (b t)

e^{,} cos (b t) t

Solution steps

\frac{d}{d b} (e^{,} sin (b t))

Treat

t

as a constant

Take the constant out:

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

= e^{,} \frac{d}{d b} (sin (b t))

Apply the chain rule:

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

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

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

= e^{,} cos (b t) t

\int csc (x) (csc (x) - cot (x)) d x

t an g e n t y = x^{3} - 2 x^{2} + 5, (2, 5)

x^{2} y^{^{'}} + 2 x y = ln (x), y (1) = 8

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

\int \frac{x + 3}{x ^{2} + 4} d x

What is the derivative of e^{,\at}sin(bt) ?
The derivative of e^{,\at}sin(bt) is e^,cos(bt)t
What is the first derivative of e^{,\at}sin(bt) ?
The first derivative of e^{,\at}sin(bt) is e^,cos(bt)t