Derivatives of inverse trig functions

  • Thread starter BuBbLeS01
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  • #1
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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?
 

Answers and Replies

  • #2
rock.freak667
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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
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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
rock.freak667
Homework Helper
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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
HallsofIvy
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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]
 

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