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I am new to mathematics somehow, and I know that:

$\frac{d(\sin x)}{dx}= \cos x$

But I can't understand this:

$\frac{d (a \sin(\omega.t))}{dt} = a.\omega\cos(\omega.t)$

where $a$ is the peak of the wave and $\omega$ is $2\pi f$ (where $f$ is frequency)

Could anyone please explain it for me?

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    What do you know about derivatives? – 2017-01-29
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    Have you seen the chain rule? – 2017-01-29
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    Just a little, I studied a small amount of information about it two years ago, and this year I saw this equation in my physics book. – 2017-01-29
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    As Sylvester said, you should apply the chain rule. It is a basic rule that you should know about derivatives. Take a time to learn about it. – 2017-01-29
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    what do you know about mathematics? tell us! – 2017-01-29
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    @PushkarSoni I studied a high school stuff about two years ago, this year I have to study physics and found that one of the topics I study is based on derivatives. And, to be honest, I don't remember so much about it. – 2017-01-29
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    good luck buddy :) – 2017-01-29
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    @PushkarSoni, Thank you, my friend! – 2017-01-29

2 Answers 2

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$\frac{d (a \sin(\omega.t))}{dt} = \frac{d(a.\sin(\omega t))}{d(\omega t)}.\frac{d(\omega t)}{dt} = a . \frac{d(\sin(\omega t))}{d(\omega t)}.\frac{\omega.d(t)}{dt}=a.\cos(\omega t).\omega$.

Here this type of above method is known as chain rule .Explore it more! Hope it helps!

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Here chain rule is used.

$\frac {d}{dt} (\sin (\omega.t)) = \cos (\omega.t). \frac {d}{dt} (\omega.t)$

= $ \cos (\omega.t). \omega. \frac{d}{dt} t$

= $ \cos (\omega.t). \omega . 1$

= $\omega. \cos (\omega.t)$