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EEJ MAIN Mathematics QUESTION #6968
Question 1

Find the derivative of $\tan^{-1}\!\left(\dfrac{\sin x-\cos x}{\sin x+\cos x}\right)$ with respect to $\dfrac{x}{2}$, for $x\in\left(0,\dfrac{\pi}{2}\right)$.

  • 1
  • $\dfrac{2}{3}$
  • $\dfrac{1}{2}$
  • 2✔️
Correct Answer Explanation

$\dfrac{\sin x-\cos x}{\sin x+\cos x} = \tan\!\left(x-\dfrac{\pi}{4}\right)$

So $f(x)=\tan^{-1}\!\left(\tan\!\left(x-\dfrac{\pi}{4}\right)\right)=x-\dfrac{\pi}{4}$ (within the principal range).

$\dfrac{df}{dx}=1$.

Let $g(x)=x/2$, so $dg/dx=1/2$.

Derivative of $f$ w.r.t. $g$: $\dfrac{df/dx}{dg/dx}=\dfrac{1}{1/2}=\mathbf{2}$