x1/6 = 6√x
The derivative of ( x1/2 ) with respect to 'x' is [ 1/2 x-1/2 ], or 1/[2sqrt(x)] .
If you mean:f(x) = x1 + root(2)The derivative of x1, or x, is simply 1. The derivative of the square root of 2, just like the derivative of any constant, is zero. Therefore, the derivative of the entire function is one.If you mean:f(x) = x1 + root(2)you shuld use the power rule (the exponent, multiplied by x to the power (exponent minus 1)):(1 + root(2)) xroot(2)If you mean:f(x) = x1 + root(2)The derivative of x1, or x, is simply 1. The derivative of the square root of 2, just like the derivative of any constant, is zero. Therefore, the derivative of the entire function is one.If you mean:f(x) = x1 + root(2)you shuld use the power rule (the exponent, multiplied by x to the power (exponent minus 1)):(1 + root(2)) xroot(2)If you mean:f(x) = x1 + root(2)The derivative of x1, or x, is simply 1. The derivative of the square root of 2, just like the derivative of any constant, is zero. Therefore, the derivative of the entire function is one.If you mean:f(x) = x1 + root(2)you shuld use the power rule (the exponent, multiplied by x to the power (exponent minus 1)):(1 + root(2)) xroot(2)If you mean:f(x) = x1 + root(2)The derivative of x1, or x, is simply 1. The derivative of the square root of 2, just like the derivative of any constant, is zero. Therefore, the derivative of the entire function is one.If you mean:f(x) = x1 + root(2)you shuld use the power rule (the exponent, multiplied by x to the power (exponent minus 1)):(1 + root(2)) xroot(2)
Invert the function.1/x = x-1x5 = 1/x-51/sqrt(x) = 1/x1/2 = x-1/2
shut up and do your hw
Assuming you mean x=1, then it is a solution.
sqr.rtx/x= sqrt.x*sqr.rtx/sqr.rtx=x/x*sqrt.x=1/sqrt.x. x1/2 = x1/2 * x1/2 = x = 1 (x1/2) /x= 1/x1/2
By factoring: x(1-x) = 0 Therefore: x = 0 or x = 1
If the variables are x1 & x2 the solution is : 1) x1=x1+x2; 2) x2=x1-x2; 3) x1=x1-x2; EX: x1=1 , x2=6; 1) x1= 1+6 = 7 2) x2= 7-6 =1 3 x1=7-1 =6 ============================================
x² = x+110 x²-x-110 = 0 x1= -(-1/2) - Square root of ((1/2)²+110) x1 = 0.5 - 10.5 x1 = - 10 x2 = -(-1/2) + Square root of ((1/2)²+110) x2 = 0.5 + 10.5 x2 = 11
-x1
no, its subtraction. if x2 is over x to the first power, the exponent at the higher value subtracts the x1. whether the x2 is above or below x1, x1 is always being subtrated from x2. X2 X(1) x __ OR __ = X or _ X(1) X2 1
x²-x-12=0 x1=-(-1/2) - Square root of ((-1/2)²+12) x1=0.5 - 3.5 x1=-3 x2=-(-1/2) + Square root of ((-1/2)²+12) x2=0.5 + 3.5 x2=4
Use: √x = x1/2 By the Power Rule (Decrease the power by 1. Multiply by the original power.): d/dx √x = d/dx x1/2 = 1/2 x-1/2
The proof of the Newton-Raphson iterative equation involves using calculus to show that the method converges to the root of a function when certain conditions are met. By using Taylor series expansion and iterating the equation, it can be shown that the method approaches the root quadratically, making it a fast and efficient algorithm for finding roots.
1 x1 x 1 = 1
x1/6 = 6√x