in Chapter 15 of Calculus Made Easy, the author proves the derivative of $ cos(\theta) $:
I'm unable to understand what did he do in the following step:
$$ dy = d(sin(\frac{\pi}{2} - \theta)) = cos(\frac{\pi}{2} - \theta)) * d(-\theta) $$
in Chapter 15 of Calculus Made Easy, the author proves the derivative of $ cos(\theta) $:
I'm unable to understand what did he do in the following step:
$$ dy = d(sin(\frac{\pi}{2} - \theta)) = cos(\frac{\pi}{2} - \theta)) * d(-\theta) $$
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, $d(\pi/2-\theta)= d(-\theta)$