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