Find the derivative of f(x)=5-|x-5|

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Find the derivative of f(x)=5-|x-5|

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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this is d/dx (5) - d/dx (x-5) derivative of 5 = 0..and the derivative of x-5 = 1 so.. 0-1 = -1
wouldnt you have a +/- value since its absolute value?
As there is a absolute value, we have to consider two cases. f(x)=5-|x-5| case i), for x>5 f(x) = 5-(x-5) = 5-x+5 =10-x f'(x) =-1 case ii) for x<=5, f(x)= 5-(-(x-5) ) =5+(x-5) =x f'(x)=1

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oh wait..that was absolute value?? so sorry! thought they were parentheses
my eyes are failing baaad
If the problem is 1/(x - 5), then, by the power rule, the derivative is: -1/(x - 5)^2 The slope, m, of the tangent line is the derivative evaluated at the x-coordinate of the point. Plug in -4 for x: m = -1/(-4 - 5)^2 = -1/81 Then solving y = mx + b for the y-intercept after plugging in the point: -1/9 = 4/81 + b b = -1/9 - 4/81 = -13/81 Tangent line equation is: y = -1/81*x - 13/81 Hope this helps.
1/(x-5) = (x-5)^-1. Then you use the chain rule to take the derivative.
|dw:1333266207662:dw| you can treat it as a piece-wise function. so if you want the derivative, f'(x) will be something different depending on the x you look at...

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