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in Chapter 15 of Calculus Made Easy, the author proves the derivative of $ cos(\theta) $:

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I'm unable to understand what did he do in the following step:

$$ dy = d(sin(\frac{\pi}{2} - \theta)) = cos(\frac{\pi}{2} - \theta)) * d(-\theta) $$

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    $\begingroup$ it's the chain rule: $\dfrac d{dx}\sin(f(x))=\cos(f(x))\dfrac d{dx}f(x)$ with $f(x)=\dfrac\pi2-\theta $ $\endgroup$ Commented Jan 12, 2020 at 0:06

2 Answers 2

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It's the chain rule: $\dfrac d{dx}\sin(f(x))=\cos(f(x))\dfrac d{dx}f(x),$ with $f(x)=\dfrac\pi2-\theta $

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You copy what the author wrote as $y= cos(\pi/2- \theta)\times d(-\theta)$ but then ask about $y= cos(\pi/2- \theta) -d(-\theta)$

Do you see that those are not the same?

Perhaps it would have easier to see if it had been written $y= cos(\pi/2- \theta)\times d(\pi/2-\theta)$. Of course since $\pi/2$ is a constant, $d(\pi/2-\theta)= d(-\theta)$

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  • $\begingroup$ The subtraction was a typing error while writing the question. Thank you, the shortcut writing $ d(-\theta) $ directly confused me. $\endgroup$
    – Jon
    Commented Jan 12, 2020 at 0:21

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