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I'm trying to find derivative of $\frac{\cos t-\sin t}{\cos t+\sin t}$ in a different way: there is a trick to find derivative of the form $\frac{ax+b}{cx+d}$: $$\left(\frac{ax+b}{cx+d}\right)'=\frac{ad-bc}{(cx+d)^2}$$ I'm trying to use it for taking derivative of $\frac{\cos t-\sin t}{\cos t+\sin t}$ too. the result of is $\frac{-2}{(\cos t+\sin t)^2}$. but I don't know how to use this method for the function

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    $\begingroup$ In addition to the answers given, you can also write it as $$(\cos(t) - \sin(t))(\cos(t) + \sin(t))^{-1}$$ In this form, you can use the product rule to find the derivative. $\endgroup$
    – Moo
    Feb 16 at 15:41
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    $\begingroup$ Also, you could differentiate $(cost+sint)y = cost-sint$ using the product rule then isolate $y'$ $\endgroup$
    – Paul
    Feb 16 at 16:05
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Your function is$$\frac{1-u}{1+u}=\frac{2}{1+u}-1$$with $u:=\tan t$, so by the chain rule its derivative is $\frac{-2\sec^2t}{(1+\tan t)^2}=\frac{-2}{(\cos t+\sin t)^2}$.

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Another approach: we can rewrite your function as $$ \frac{\cos t - \sin t}{\cos t + \sin t} = \frac{\cos(t)\sin(\pi/4) - \sin(t)\cos(\pi/4)}{\cos(t)\cos(\pi/4) + \sin(t) \sin(\pi/4)} = \frac{\sin(\pi/4 - t)}{\cos(\pi/4 - t)} = \tan(\pi/4 - t). $$ By the chain rule, we find the derivative of this function to be $$ -\sec^2(\pi/4 - t) = -\frac{1}{\cos^2(\pi/4 - t)} = -\frac{1}{[\cos(t)/\sqrt{2} + \sin(t)/\sqrt{2}]^2} \\= \frac{-2}{(\cos(t)+\sin(t))^2}. $$

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  • $\begingroup$ Did you lose a negative sign? $\endgroup$
    – Moo
    Feb 16 at 16:03
  • $\begingroup$ @BenGrossman You definitely lost a negative sign on the numerator. Should be $-\tan(t-\frac{\pi} 4)$. $\endgroup$
    – Deepak
    Feb 16 at 16:14
  • $\begingroup$ @Deepak Thanks, fixed it $\endgroup$ Feb 16 at 16:17
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Instead of trying to get the derivative of $f=\frac{\cos-\sin}{\cos+\sin}$, try to get the derivative of $f^2$. This way you have that $$f^2(t)=\dfrac{-2\sin(t)\cos(t)+1}{2\sin(t)\cos(t)+1}.$$ Now you can apply the trick by taking $a=-2$, $b=d=1$, $c=2$ and $x=\sin(t)\cos(t)$ in order to get $(f^2)'$. Then just apply the chain rule and solve for $f'$.

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You can also do: $$\frac{\cos{t}-\sin{t}}{\cos{t}+\sin{t}} = \frac{(\cos{t}-\sin{t})(\cos{t}-\sin{t})}{(\cos{t}+\sin{t})(\cos{t}-\sin{t})} = \frac{1-\sin{2t}}{\cos^2{t}-\sin^2{t}} = \frac{1}{\cos{2t}}-\tan{2t} = \sec{2t}-\tan{2t}$$ Take the derivative to get $$2\sec{2t}(\tan{2t}-\sec{2t}) = \frac{-2}{(\cos{t}+\sin{t})^2}$$ after some manipulation.

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Inspired by J.G.'s answer we can rewrite $\frac{\cos t-\sin t}{\cos t+ \sin t}$ as $\frac{1-\tan t}{1+\tan t}$. Let's call it $f(t)$:

$$f(t)=\frac{-\tan t+1}{\tan t+1}$$ Then by chain rule and by using the trick I mentioned in my question we have : $$\frac{df}{dt}=\frac{df}{d(\tan t)}\times\frac{d(\tan t)}{dt}=\frac{-2}{(1+\tan t)^2}\times\sec^2 t=\frac{-2}{(\cos t+\sin t)^2}$$

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