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Problem:

Differentiate with respect to $x$: $\arctan(\frac{\cos x-\sin x}{\cos x+\sin x})$

My attempt:

Let, $y=\arctan(\frac{\cos x-\sin x}{\cos x+\sin x})$

$$y=\arctan(\frac{\cos x-\sin x}{\cos x+\sin x})$$

$$y=\arctan\frac{(\cos x-\sin x)^2}{2\cos2x}$$

$$y=\arctan\frac{1-\sin2x}{2\cos2x}$$

$$y=\arctan(\frac{1}{2\cos2x}-\frac{\sin2x}{2\cos2x})$$

$$y=\arctan\frac{1}{2}(\frac{1}{\cos2x}-\frac{\sin2x}{\cos2x})$$

$$y=\arctan\frac{1}{2}(\sec2x-\tan2x)$$

My observations:

Now, I could find the derivative using the brute force of chain rule, but the derivative of the above graph is $-1$, so I think a much easier way to find the derivative might exist.

Question:

  1. Is there a way to find the derivative of the above graph very easily, which is not that tedious?
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  • $\begingroup$ Do you now how to caclculate the derivative of inverse functions? $\endgroup$ Oct 21 '21 at 8:07
  • $\begingroup$ yes, $\frac{d}{dx}\arctan(x)=\frac{1}{1+x^2}$ $\endgroup$ Oct 21 '21 at 8:08
  • $\begingroup$ Btw how did the factor 2 appeared in you denominator? $\endgroup$ Oct 21 '21 at 8:08
  • $\begingroup$ If you know the derivative of arctan, then you can simply use the chain rule. $\endgroup$ Oct 21 '21 at 8:10
  • $\begingroup$ @GáborPálovics I edited the question; I know, but it seems tedious. The derivative of the above graph is a constant (-1), so I thought that there might be a really easy way to find the derivative then. $\endgroup$ Oct 21 '21 at 8:13
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We have $$\begin{align*}\frac{\cos x-\sin x}{\cos x+\sin x}&=\frac{\frac{\cos x-\sin x}{\sqrt2}}{\frac{\cos x+\sin x}{\sqrt2}} \\&=\frac{\sin{(\frac{\pi}4}-x)}{\cos{(\frac{\pi}4}-x)} \\&=\tan{(\frac{\pi}4}-x)\end{align*}$$ Can you proceed form here?

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