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Find the Equation of the Tangent Line


Date: 1/21/96 at 23:52:54
From: S. W. Davis 
Subject: Equation of the tangent line

Find the the equation of the tangent line when x=1

f(x)= a/b

a=2^x+1-sin(x+1/X+2)

b=(x^3-2x+7)^1/3


Date: 7/13/96 at 22:43:43
From: Doctor Paul
Subject: Re: Equation of the tangent line

        2^x + 1 - sin((x+1)/(x+2)
f(x)=   -------------------------
            (x^3-2x+7)^1/3

The first thing you must know is how to find the equation of a 
line:

   (y-y0) = m(x-x0)

where m is the slope and (x0,y0) is a point on the line.

We know what x0 is; that's given in the problem: it's 1.
If we plug 1 in for x then we can get a number for f(x) and that 
will be out y0.  That's not so hard.  Then all we need is m. 

Let's first get y0.

        2^1 + 1 - sin(2/3)
f(x) =  ------------------
          (1^3-2(1)+7)^(1/3)

         3 - .6183698031
     =   ---------------
           1.817120593 

     =   1.310661607

so our formula looks like this:

(y-1.310661607) = m*(x-1)
     
To find m, we take the derivitive of f(x), and plug in x0 and y0

This derivitive is extremely messy..I used the quotient rule and 
here's what I got (broken up into two pieces..) :

 x             x + 1  /  1       x + 1 \
2  ln(2) - cos(-----) |----- - --------|      
               x + 2  |x + 2          2|       
                      \        (x + 2) /    
 --------------------------------         - 
              3           1/3                           
            (x  - 2 x + 7)      

      / x           x + 1 \     2
      |2  + 1 - sin(-----)| (3 x  - 2)
      \             x + 2 /                     
      --------------------------------
             3          4/3
          3*(x  - 2x + 7)

now plug in x = 1 and get the slope to be .6420381621

so the equation of the line is:

(y-1.310661607) = .6420381621*(x-1)

-Doctor Paul,  The Math Forum

    
Associated Topics:
Middle School Equations

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