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

In summary, the derivative of \arccos (4x^2) is \frac{dy}{dx} = -\frac{8x}{\sqrt{1 - 16x^4}} and the domain for which the derivative applies is x\ \epsilon\ (-\frac{1}{4},\ \frac{1}{4}). The mistake in line 3 is that the argument of arccos should be 4x^2 instead of just x.
  • #1
ultima9999
43
0
Yeah, I tried doing this question and just wanted to check if I was correct.

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

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

[tex]\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*}[/tex]

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

[tex]\therefore \frac{dy}{dx} = \frac{4\sqrt{\cos y}}{-\sqrt{1 - \cos^2 y}}[/tex]
 
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  • #2
Don't forget that cos y= 4x2 so
[tex]\frac{dy}{dx}= -\frac{8x}{\sqrt{1- 16x^2}}[/tex]
I suspect your teacher will prefer that form.
 
  • #3
Thanks, and I think it should be
[tex]\frac{dy}{dx} = -\frac{8x}{\sqrt{1 - 16x^4}}[/tex]
 
  • #4
ultima9999 said:
Thanks, and I think it should be
[tex]\frac{dy}{dx} = -\frac{8x}{\sqrt{1 - 16x^4}}[/tex]

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
George Jones said:
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 [tex]x\ \epsilon\ (-\frac{1}{4},\ \frac{1}{4})[/tex] simply because when x = 0.25, the denom is undefined.

And what is the mistake in line 3?
 
  • #6
ultima9999 said:
The domain would be [tex]x\ \epsilon\ (-\frac{1}{4},\ \frac{1}{4})[/tex] 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 [itex]\left[ -1 , 1 \right][/itex], i.e., [itex]4x^{2} \epsilon \left[ -1 , 1 \right][/itex], not [itex]x[/itex]. This immediately gives "almost" the domain of validity.
 
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Related to What is the mistake in finding the derivative of arccos (4x^2)?

1. What is the inverse cosine function?

The inverse cosine function, also known as the arccosine function, is the inverse of the cosine function. It is denoted as cos-1 or arccos and is used to find the angle whose cosine is a given number.

2. What is the derivative of the inverse cosine function?

The derivative of the inverse cosine function, or cos-1, is equal to -1/sqrt(1-x2). This can also be written as -1/√(1-x2).

3. How do you find the derivative of the inverse cosine function?

To find the derivative of the inverse cosine function, you can use the chain rule. First, rewrite cos-1 as arccos. Then, take the derivative of arccos, which is -1/sqrt(1-x2). Finally, use the chain rule to multiply by the derivative of the inner function, which is simply 1. This results in a final derivative of -1/sqrt(1-x2).

4. What is the domain of the inverse cosine function?

The domain of the inverse cosine function, or cos-1, is -1 to 1. This means that the input or x-values must be between -1 and 1 for the function to be defined.

5. How is the inverse cosine function used in real life?

The inverse cosine function is used in many real-life applications, such as in physics and engineering, to solve problems involving angles. It is also used in trigonometry and calculus to find the inverse of a given cosine value. Additionally, it is used in computer graphics and programming to calculate the angle of rotation needed for objects to move on a curved path.

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