# Homework Help: Inverse Cos Derivative

1. Jun 27, 2006

### ultima9999

Yeah, I tried doing this question and just wanted to check if I was correct.

Write down the derivative of $$\arccos (4x^2)$$ and state the domain for which the derivative applies.

$$\mbox{let}\ y = \arccos (4x^2), x\ \epsilon\ [-1, 1], y\ \epsilon\ [0, \pi]$$
$$\Leftrightarrow x = \frac{\sqrt{\cos y}}{2}$$

\begin{align*} \frac{d}{dx} x = \frac{d}{dx} \frac{\sqrt{\cos y}}{2} \\ 1 = \frac{-\sin y}{4 \sqrt{\cos y}} \cdot \frac{dy}{dx} \\ \frac{dy}{dx} = \frac{4 \sqrt{\cos y}}{-\sin y} \end{align*}

$$\cos^2 y + \sin^2 y = 1$$
$$\sin y = \sqrt{1 - \cos^2 y}, \mbox{because}\ y\ \epsilon\ [0, \pi]\ \mbox{so}\ \sin y \geq 0$$

$$\therefore \frac{dy}{dx} = \frac{4\sqrt{\cos y}}{-\sqrt{1 - \cos^2 y}}$$

Last edited: Jun 27, 2006
2. Jun 27, 2006

### HallsofIvy

Don't forget that cos y= 4x2 so
$$\frac{dy}{dx}= -\frac{8x}{\sqrt{1- 16x^2}}$$
I suspect your teacher will prefer that form.

3. Jun 27, 2006

### ultima9999

Thanks, and I think it should be
$$\frac{dy}{dx} = -\frac{8x}{\sqrt{1 - 16x^4}}$$

4. Jun 27, 2006

### George Jones

Staff Emeritus
What about the domain of validity. Note that with with respect to this, there is a mistake in line 3 (first line of mathematics) in post #1.

5. Jun 27, 2006

### ultima9999

The domain would be $$x\ \epsilon\ (-\frac{1}{4},\ \frac{1}{4})$$ simply because when x = 0.25, the denom is undefined.

And what is the mistake in line 3?

6. Jun 27, 2006

### George Jones

Staff Emeritus
The argument of arccos is an element of the closed interval $\left[ -1 , 1 \right]$, i.e., $4x^{2} \epsilon \left[ -1 , 1 \right]$, not $x$. This immediately gives "almost" the domain of validity.

Last edited: Jun 27, 2006