+322 Find an equation of the tangent line to the curve at the point (2,2).
\(y=\frac{4x}{x+2}\)
+129852 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
