Derivatives of inverse trig functions

In summary, the conversation discusses finding the equation of a tangent line to the graph of the function f(x) = arcsin2x at a given point. The solution involves using the formula \frac{d}{dx}(sin^{-1}X)=\frac{1}{\sqrt{1-X^2}} and the chain rule. The conversation also mentions a more general formula for differentiating arcsine functions and clarifies that it is equivalent to the initial formula given.
  • #1
BuBbLeS01
602
0

Homework Statement


find an equation of the tangent line to graph of f at the indicated point.

f(x) = arcsin2x
(u'/sqrt 1-u^2)(2)


2. The attempt at a solution
(1/sqrt 1-2x^2)(2)
I got the answer from calcchat but I don't understand where the 1 and 2 came from?
 
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  • #2
Let [tex]y=sin^{-1}2x=> siny=2x[/tex] can you do implicity Differentiation?

or here is a formula to know

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

and because you know that formula...you can use the chain rule
 
  • #3
Okay I thought the formula was...
u'/sqrt 1-u^2
so can u bring it up so its 1-(1-x)^-1/2?
 
  • #4
BuBbLeS01 said:
Okay I thought the formula was...
u'/sqrt 1-u^2
so can u bring it up so its 1-(1-x)^-1/2?

[tex]\frac{1}{\sqrt{1-X^2}} ={1}{(1-X^2)^{\frac{1}{2}} =(1-X^2)^{\frac{-1}{2}}[/tex]

because [tex]\frac{a}{b^n} = a*b^{-n}[/tex]
 
  • #5
BuBbLeS01 said:
Okay I thought the formula was...
u'/sqrt 1-u^2
so can u bring it up so its 1-(1-x)^-1/2?

Obviously, if u= x (and the differentiation is with respect to x) then u'= 1 so the two formulas are the same. The formula you give is more general. If u is any function of x (and the differentiation is with respect to x) then
[tex]\frac{d arcsin(u(x))}{dx}= \frac{ \frac{du}{dx}}{\sqrt{1- u^2(x)}}[/tex]
 

1. What are the derivatives of inverse trig functions?

The derivatives of inverse trig functions are the inverse sine, inverse cosine, and inverse tangent functions. They are denoted as sin^-1(x), cos^-1(x), and tan^-1(x), respectively. The derivatives of these functions are cos(x), -sin(x), and 1/(1+x^2), respectively.

2. How do you find the derivative of inverse trig functions?

To find the derivative of inverse trig functions, you can use the chain rule. For example, to find the derivative of inverse sine (sin^-1(x)), you would first rewrite it as arcsin(x) and then use the chain rule to get 1/sqrt(1-x^2).

3. What is the domain of inverse trig functions?

The domain of inverse trig functions is usually restricted to specific intervals in order to make them one-to-one functions. For example, the domain of inverse sine (sin^-1(x)) is [-1,1], while the domain of inverse tangent (tan^-1(x)) is all real numbers.

4. What is the relationship between inverse trig functions and their derivatives?

The derivative of an inverse trig function is the inverse of the derivative of the corresponding trig function. This means that the derivative of inverse sine (sin^-1(x)) is equal to 1/cos(x), the derivative of inverse cosine (cos^-1(x)) is equal to -1/sin(x), and the derivative of inverse tangent (tan^-1(x)) is equal to 1/(1+x^2).

5. How are inverse trig functions used in calculus?

Inverse trig functions are used to solve integrals and to find the slopes of curves at specific points. They are also used in solving real-world problems that involve angles or distances, such as in physics or engineering. Inverse trig functions are also important in the study of trigonometric identities and equations.

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