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Derive with respect to $\arccos\left(x^2\right)$ where $$f(x)=\arctan\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right)$$

Can someone please explain to me what does "with respect to $\arccos\left(x^2\right)$" mean ?

Blue
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Mo2edmatic
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  • By the chain rule,$$\frac{df}{d\arccos x^2}=\frac{dx}{d\arccos x^2}\frac{df}{dx}=\frac{\frac{df}{dx}}{\frac{d\arccos x^2}{dx}}$$for a differentiable function $f(x)$. – J.G. Jun 18 '22 at 20:43
  • Assuming that you did not mean to say differentiate, and instead intended derive, it means that you let $\theta$ equal the angle such that $\cos(\theta) = x^2.$ Then, the assignment is to re-express $f(x)$ in terms of the variable $(\theta)$, rather than the variable $(x)$. In other words, the assignment is asking you to construct the function $g(\theta)$ such that $g(\theta) = f(x).$ – user2661923 Jun 18 '22 at 21:11
  • Re last comment, as a starting point, you have that $$\sqrt{1 + x^2} = \sqrt{1 + \cos(\theta)}.$$ – user2661923 Jun 18 '22 at 21:14

2 Answers2

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I believe the question wants you to substitute $y=\arccos(x^2)$ and find $f(x)$ in terms of $y$.

So,

$$ \begin{align} f(x)&=\arctan\left( \frac{\sqrt{1+\cos y} - \sqrt{1-\cos y}} {\sqrt{1+\cos y} + \sqrt{1-\cos y}}\right)\\ &=\arctan\left( \frac{\cos(y/2) - \sin(y/2)} {\cos(y/2) + \sin(y/2)}\right)\\ &=\arctan\left( \frac{\cos(y/2+\pi/4)} {\sin(y/2+\pi/4)}\right)\\ &=\arctan(\cot(y/2+\pi/4))\\ &=\arctan(\tan(\pi/4-y/2))\\ &=\frac{\pi}{4}-\frac{y}{2}\\ &=\frac{\pi}{4}-\frac{\arccos(x^2)}{2} \end{align} $$

Indeed Wolfram Alpha agrees (up to computational error).

Even if the question wants you to compute the derivative, it should be pretty easy from the last line above.

I used half-angle formulae, sum and difference identities, and reflection between cot and tan.

Benjamin Wang
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If you meant to say "Derivative", it means $$ \frac{df(x)}{d \arccos(x^{2})}. $$ Please see my answer here.

Artur Wiadrowski
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