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I had a question regarding the differentiation of trigonometric functions.

To be a bit more specific, I'm currently studying a chapter about vector functions and vector calculus for an engineering mathematics course I'm taking at school. There is a rather simple detail that I don't understand though for an example question in the textbook.

In the particular example question, the vector function is given as:

$$\mathbf{r}(t) = \cos 2t\ \mathbf{i} + \sin t\ \mathbf{j}$$

And the differentiation of the function is apparently as follows:

$$\mathbf{r^\prime}(t) = -2\sin 2t\ \mathbf{i} + \cos t\ \mathbf{j}$$

What I'm having trouble understanding is why the $-2$ has been added to the front of the $\sin$ for the differential equation? I thought that

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

Thank you.

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Recall that by chain rule

$$(\cos f(t))'=-\sin f(t) \cdot f'(t)$$

and in this case we have $f(t)=2t\implies f'(t)=2$.

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  • $\begingroup$ Thank you very much. I don't know why I didn't think of the chain rule. $\endgroup$ – Seankala Jun 11 '18 at 10:28
  • $\begingroup$ @Sean You are welcome! I'm sure that next time you'll keep it in mind! Bye $\endgroup$ – gimusi Jun 11 '18 at 10:28

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