Why Is \cos y Positive When Differentiating \arcsin x?

In summary, when differentiating \arcsin x, we use the relation x=\sin y \implies \frac{dx}{dy}= \cos y and \sin^2 y + \cos^2 y = 1 to find that \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}. The reason why \cos y = \sqrt{1 - \sin^2 y} instead of \cos y = -\sqrt{1 - \sin^2 y} is because y = \arcsin x is always in the range [-\pi/2, \pi/2], ensuring that \cos y must be positive.
  • #1
perishingtardi
21
1
Say we want to differentiate [itex]\arcsin x[/itex]. To do this we put [tex]y=\arcsin x.[/tex] Then [tex]x=\sin y \implies \frac{dx}{dy}= \cos y.[/tex] Then we use the relation [tex]\sin^2 y + \cos^2 y = 1 \implies \cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}.[/tex] Therefore [tex]\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}.[/tex]

My question is that when we use [tex]\sin^2 y + \cos^2 y = 1,[/tex] how do we know that [tex]\cos y = \sqrt{1 - \sin^2 y}[/tex] rather than [tex]\cos y = -\sqrt{1 - \sin^2 y}[/tex]?
 
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  • #2
Because arcsine, by convention, always returns angles in the range [itex][-\pi/2, \pi/2][/itex]. This guarantees that [itex]y = \arcsin x[/itex] is in that range, so [itex]\cos y[/itex] has to be positive. Therefore, you know it must be the positive square root, instead of the negative one.
 

Related to Why Is \cos y Positive When Differentiating \arcsin x?

1. What is the definition of inverse trigonometric functions?

The inverse trigonometric functions, also known as arc trigonometric functions, are the inverse functions of the trigonometric functions (sin, cos, tan, cot, sec, csc). They give the angle whose trigonometric ratio is a given value.

2. What is the process of differentiating inverse trigonometric functions?

The process of differentiating inverse trigonometric functions involves using the chain rule and the derivatives of the inverse trigonometric functions. For example, the derivative of arctan(x) is 1/(1+x^2) and the derivative of arcsin(x) is 1/sqrt(1-x^2).

3. What are the common derivative formulas for inverse trigonometric functions?

The common derivative formulas for inverse trigonometric functions are:- d/dx(arcsin(x)) = 1/sqrt(1-x^2)- d/dx(arccos(x)) = -1/sqrt(1-x^2)- d/dx(arctan(x)) = 1/(1+x^2)- d/dx(arccot(x)) = -1/(1+x^2)- d/dx(arcsec(x)) = 1/(x*sqrt(x^2-1))- d/dx(arccsc(x)) = -1/(x*sqrt(x^2-1))

4. What are the key points to remember when differentiating inverse trigonometric functions?

The key points to remember when differentiating inverse trigonometric functions are:- Use the chain rule- Remember the derivatives of the inverse trigonometric functions- Pay attention to the signs (positive or negative) in the derivatives- Simplify the final answer as much as possible

5. Are there any real-life applications of differentiating inverse trigonometric functions?

Yes, there are many real-life applications of differentiating inverse trigonometric functions, such as:- In physics: for calculating the velocity, acceleration, and displacement of objects in motion- In engineering: for designing and analyzing mechanical and electrical systems- In navigation: for calculating the trajectories and angles of ships and planes- In astronomy: for predicting and analyzing the movements of celestial bodies- In finance: for modeling and predicting stock market trends

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