f(x)=1
f'(x)=0 because the derivative of a constant is ALWAYS 0.
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The original equation f(x) = 6/(x+3) can be rewritten as f(x) = 6(x+3)-1. Now derive the equation according the the power rule and the chain rule: y = 6 (x+3)-1 dy/dx = 6 (-1)(x+3)-2(1)* dy/dx = -6/(x+3)2 * by the chain rule, you must multiply by the derivative of which is simply one. Thus, the derivative of f(x) = 6/(x+3) equals -6/(x+3)2
14x
x= (x+1)/x(x+1)/x -x = 0f'(x) = ((x*1)-(x+1)) /x^2 -11/x^2 - 1
The derivative of f(x) = x mod b is f'(x)=1, except where x is a multiple of b, when it is undefined.