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The function $\cos^{-1}(x)$ is defined for $-1\leq x\leq1$ and $0\leq y\leq \pi$ by the equivalence

$$y=\cos^{-1}(x)\Leftrightarrow x=\cos y$$

(a) Compute $\dfrac{dy}{dx}$ as a function of $y$ by differentiating the second equation in $(1)$.

(b) Use the identity $\sin y=\sqrt{1-\cos^2 y}$ and make the substitution $x=\cos y$ in (a) to prove that

$$\dfrac{d}{dx}\cos^{-1}(x)=\frac{-1}{\sqrt{1-x^2}}$$

rae306
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Inverse Problem
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  • better to write the question, not to attach a low quality picture – chouaib Oct 01 '14 at 06:29
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    Are you aware that $dy/dx = 1/(dx/dy)$? Do you know how to differentiate $x = \cos y$ with respect to $y$? Combining these two things is enough to solve the problem. – Robin Saunders Oct 01 '14 at 06:30

1 Answers1

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DISCLAIMER: See this answer only after you have attempted to solve the problem yourself.

I am providing an answer for future reference. First differentiate both sides and utilize the chain rule.

$$\begin{align} \frac{d}{dx}(x)&=\frac{dy}{dx}\frac{d}{dy}(\cos(y))\\ 1&=\frac{dy}{dx}(-\sin(y)) \end{align} $$ Dividing both sides by $-\sin(y)$ we obtain. $$ \begin{align} \frac{dy}{dx}&=\frac{-1}{\sin(y)}\\ &=\frac{-1}{\sqrt{1-x^2}} \end{align} $$

In the last part we used the substitution $\sin(y)=\sqrt{1-cos^2(y)}=\sqrt{1-x^2}$. Furthermore, since $\frac{dy}{dx}=\frac{d}{dx}cos^{-1}(x)$ we have established the fact that $$\frac{d}{dx}cos^{-1}(x)=\frac{-1}{\sqrt{1-x^2}}$$

Enigma
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