Understanding d/dx (sin-1x) & Deriving dy/dx = -sin-2xcosx

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So the notation is one letter short of being confusing.In summary, the derivative of sin-1x is 1/sqrt(1-x^2) and not 1/sin x. This is because sin-1x represents the inverse function of sin x, not the reciprocal. This can be shown using the chain rule, which yields -sin-2x cosx as the derivative of sin-1x.
  • #1
Kyoma
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d/dx (sin-1x) = [itex]\frac{1}{\sqrt{1-x2}}[/itex]

But using chain rule, I got:

dy/dx = -sin-2xcosx

Why?
 
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  • #2
Kyoma said:
d/dx (sin-1x) = [itex]\frac{1}{\sqrt{1-x2}}[/itex]

But using chain rule, I got:

dy/dx = -sin-2xcosx

Why?

sin-1x is not equal to [itex]\frac{1}{\sin x}[/itex]. sin-1x stands for the arcsin of x, or more specifically, the angle y that satisfies sin y = x.

I've always personally disliked the notation sin-1x, but it is what it is.
 
  • #3
If [itex]y= sin^{-1}(x)[/itex] then [itex]x= sin(y)[/itex].

[tex]\frac{dx}{dy}= cos(y)[/tex]

so that
[tex]\frac{dy}{dx}= \frac{1}{cos(y)}[/tex]

but [tex]y= sin^{-1}(x)[/tex] so [tex]cos(sin^{-1}(x))= \sqrt{1- sin^2(sin^{-1}(x))}= \sqrt{1- x^2}[/tex]

so
[tex]\frac{d(sin^{-1}(x)}{dx}= \frac{1}{\sqrt{1- x^2}}[/tex]

As gb7nash said, the "-1", applied to functions denotes the "inverse function" (the inverse for function composition), not the reciprocal (the inverse for multiplication).
 

1. What is the definition of d/dx (sin-1x)?

The notation d/dx represents the derivative of a function with respect to the independent variable x. In the case of sin-1x, it represents the derivative of the inverse sine function with respect to x.

2. How do you derive the equation dy/dx = -sin-2xcosx?

To derive this equation, we can use the chain rule and the derivative of the inverse trigonometric function. First, we rewrite sin-1x as arcsin(x). Then, using the chain rule, we get dy/dx = d/dx(sin(arcsin(x))). By applying the derivative of the inverse trigonometric function, we get dy/dx = cos(arcsin(x)). Finally, using the Pythagorean identity sin2x + cos2x = 1, we can rewrite cos(arcsin(x)) as √(1-x2), giving us dy/dx = √(1-x2). Since sin-1x represents the inverse of sinx, we can replace sinx with x to get dy/dx = √(1-x2) = √(1-x2) * 1 = √(1-x2) * sin-2x, which simplifies to dy/dx = -sin-2xcosx.

3. What is the domain and range of sin-1x?

The domain of sin-1x is [-1, 1], which means that the input value x can range from -1 to 1. The range of sin-1x is [-π/2, π/2], which means that the output value of arcsin(x) can range from -π/2 to π/2. This is because the inverse sine function only returns values within this range.

4. How does the graph of sin-1x compare to the graph of sinx?

The graph of sin-1x is the inverse of the graph of sinx. This means that the graph of sin-1x is a reflection of the graph of sinx over the line y=x. The graph of sin-1x is also a curve that increases from -π/2 to π/2, similar to the graph of sinx.

5. How can d/dx (sin-1x) be applied in real-life situations?

The derivative of sin-1x can be used to solve problems in physics, engineering, and other fields that involve the inverse sine function. For example, it can be used to find the slope of a curve, which is critical in determining the velocity and acceleration of an object in motion. It can also be used in calculating the rate of change of various physical quantities, such as temperature or pressure, in different scenarios.

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