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

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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?
This step is incorrect: (y+5/16)=45/32(x-2) (y+5/16)=45/32x-19/32 |dw:1349146268366:dw|

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Other answers:

45/32x-19/32... that's what i keep getting
haha ohh.. stupid me
=) great job otherwise!!! you did all the hard stuff perfectly! +fan
yay! thank you!

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