Question

Suppose that f(x)=((5x)/((2-2x)^5) Find an equation for the tangent line to the graph of at x=2.

Answer 1

y= -5/16
x= 2

y'=((5(4x+1))/((32(x-1)^6)))
y'(2)=45/32
45/32=slope

(y-(-5/16))=45/32(x-2)
(y+5/16)=45/32(x-2)
(y+5/16)=45/32x-19/32
y=45/32x-29/32


this is what I came up with and it's wrong... Can someone help me?

Answer 2

This step is incorrect:
(y+5/16)=45/32(x-2)
(y+5/16)=45/32x-19/32