What is the mistake in finding the derivative of arccos (4x^2)?

[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*}<br /> \frac{d}{dx} x = \frac{d}{dx} \frac{\sqrt{\cos y}}{2} \\<br /> 1 = \frac{-\sin y}{4 \sqrt{\cos y}} \cdot \frac{dy}{dx} \\<br /> \frac{dy}{dx} = \frac{4 \sqrt{\cos y}}{-\sin y}<br /> \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}}

 

[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.

 

[ultima9999]

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

 

[George Jones]

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

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.

 

[ultima9999]

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.

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?

 

[George Jones]

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?

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.