Derivative of 4x is 4.
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X^4 ? 4x^3
4/x can be written as 4x-1 (the power of negative 1 means it is the denominator of the fraction) 4*-1 = -4 Therefore, the derivative is -4x-2
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To find the extreme value of the parabola y = x2 - 4x + 3 ...(1) Take the derivative of the equation.y = x2 - 4x + 3y' = 2x - 4(2) Set the derivative = 0 and solve for x.y' = 2x - 40 = 2x - 42x = 4x = 4/2x = 2(3) Plug this x value back into the original equation to find the associated y coordinate.x = 2y = x2 - 4x + 3y = (2)2 - 4(2) + 3y = 4 - 8 + 3y = -1So the vertex is at (2, -1).
y=(8x).5 + (4x).5 = (2+2sqrt(2))x.5 y'=(1 + sqrt(2))/sqrt(x)