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## I think of two versions of

I think of two versions of that problem.

1.) I got to get the derivative of $\sin(at)$ first before getting the Laplace transform of the derivative of $\sin(at).$

2.) This form: $\mathcal L\{ \frac{d}{dt}(sin (at)) \}$

I think I go do the #1.......because there is another operation that deas with Laplace transforms of derivatives.

## Well...to get the derivative

To get the Laplace transform of derivative of $\sin (at)$, get first the derivative of $\sin (at)$. So we let $u = at$ and $du = a.$ Then recall that $\frac{d}{dx}(\sin u) = \cos(u) du$. So....

$$\frac{d}{dx}(\sin u ) = \cos (u) du$$ $$\frac{d}{dx}(\sin (at) ) = \cos (at) (a)$$ $$\frac{d}{dx}(\sin (at) ) = a \cos (at) $$

We will get the Laplace transform of $a \cos (at)$.

Recall that the Laplace transform of $\cos (\omega_o t)$ is $\frac{s}{s^2 + \omega_o ^2}$. Then the Laplace transform of $a \cos (at)$ would be:

$$\mathcal L \{ \cos (\omega_o t) \} = \frac{s}{s^2 + \omega_o ^2}$$ $$\mathcal L \{ a \cos (at) \} = a \left( \frac{s}{s^2 + (a)^2} \right)$$ $$\mathcal L \{ a \cos (at) \} = a \left( \frac{s}{s^2 + a^2} \right)$$

We conclude that Laplace transform of $a \cos (at)$ is $ \frac{a s}{s^2 + a^2} $.

Hope it helps:-)

## hi Lee:-)

In reply to Well...to get the derivative by fitzmerl duron

hi Lee:-)