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Find an equation of the tangent line to the curve at the point (2,2).

 

\(y=\frac{4x}{x+2}\)

 Feb 18, 2019
 #1
avatar+102434 
+2

We have

 

y = (4x) ( x + 2)^(-1)        the derivative of this gives the slope of the graph at any point x

 

So....the derivative is

 

y ' (x)  =  4(x + 2)^(-1) - (4x)(x +  2)^(-2)

 

So...the slope at  x = 2 is given by

 

y ' (2)   =   2(2 + 2) ^(-1) -  (4*2)(2 + 2)^(-2)  =   1 - (1/2) = 1/2

 

So....using the slope and the point (2,2).....the equation of the tangent line is

 

y = (1/2) (x - 2) + 2

 

y  = (1/2)x - 1 + 2

 

y = (1/2)x + 1

 

Here's a graph : https://www.desmos.com/calculator/a65are5fgn

 

P.S......if you need to take the derivative using the difference quotient....let me know

 

cool  cool cool

 Feb 18, 2019
 #2
avatar+322 
0

I don't.  I just learned the very easy way of doing it.  It's pretty nice.

Ruublrr  Feb 18, 2019
 #3
avatar+102434 
+2

You're not kidding....!!!

 

If I had to use that Difference Quotient everytime i had to take a derivative....I'd have never made it through Calc I

 

LOL!!!!

 

 

cool cool cool

CPhill  Feb 18, 2019

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