x*x1/2= x3/2 Derivative = 3/2 * x1/2
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)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)
total differentiation is closer to implicit differentiation although you are not solving for dy/dx. in other words: the total derivative of f(x1,x2,...,xk) with respect to xn= [df(x1,x2,...,xk)/dx1][dx1/dxn] + df(x1,x2,...,xk)/dx2[dx2/dxn]+...+df(x1,x2,...,xk)/dxn +[df(x1,x2,...,xk)/dxn+1][dxn+1/dxn]+...+[df(x1,x2,...,xk)/dxk][dxk/dxn] however, the partial derivative is not this way. the partial derivative of f(x1,x2,...,xk) with respect to xn is just that, can't be expanded. The chain rule is not the same as total differentiation either. The chain rule is for partially differentiating f(x1,x2,...,xk) with respect to a variable not included in the explicit form. In other words, xn has to be considered a function of this variable for all integers n. so the total derivative is similar to the chain rule, but not the same.
The derivative of ( x1/2 ) with respect to 'x' is [ 1/2 x-1/2 ], or 1/[2sqrt(x)] .
The derivative of a function is another function that represents the slope of the first function, slope being the limit of delta y over delta x at any two points x1,y1 and x2,y2 on the graph of the function as delta x approaches zero.
x*x1/2= x3/2 Derivative = 3/2 * x1/2
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)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)
total differentiation is closer to implicit differentiation although you are not solving for dy/dx. in other words: the total derivative of f(x1,x2,...,xk) with respect to xn= [df(x1,x2,...,xk)/dx1][dx1/dxn] + df(x1,x2,...,xk)/dx2[dx2/dxn]+...+df(x1,x2,...,xk)/dxn +[df(x1,x2,...,xk)/dxn+1][dxn+1/dxn]+...+[df(x1,x2,...,xk)/dxk][dxk/dxn] however, the partial derivative is not this way. the partial derivative of f(x1,x2,...,xk) with respect to xn is just that, can't be expanded. The chain rule is not the same as total differentiation either. The chain rule is for partially differentiating f(x1,x2,...,xk) with respect to a variable not included in the explicit form. In other words, xn has to be considered a function of this variable for all integers n. so the total derivative is similar to the chain rule, but not the same.
Write square root of x as x1/2. Then use the formula for the derivative of a power.
By the chain rule, the derivative of sin(x1/2) will be the derivative of x1/2 multiplied by the derivative of the enclosing sine function. Thus, y = sin(x1/2) y' = (1/2)*(x-1/2)*cos(x1/2) For further reading, you might want to consult your calculus book on the chain rule. Here is a site that (kind of) explains the chain rule, though it does have good examples: http://archives.math.utk.edu/visual.calculus/2/chain_rule.4/index.html For step-by-step derivatives of functions, try Calc 101: http://calc101.com/webMathematica/derivatives.jsp
The derivative of ( x1/2 ) with respect to 'x' is [ 1/2 x-1/2 ], or 1/[2sqrt(x)] .
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
The derivative of a function is another function that represents the slope of the first function, slope being the limit of delta y over delta x at any two points x1,y1 and x2,y2 on the graph of the function as delta x approaches zero.
Aaron Poison/Bug x1 Bug/Fighting x1 Bug/Flying x2 Poison/Dark x1 Bertha Water/Ground x2 Ground x1 Rock x1 Rock/Ground x1 Flint Fire x1 Fire/Fighting x1 Steel/Ground x1 Normal x1 Ghost/Flying x1 Lucian Psychic x2 Normal/Psychic x1 Fight/Psychic x1 Steel/Psychic x1 Cynthia Ghost/Dark x1 Dragon/Ground x1 Water/Ground x1 Water x1 Grass/Poison x1 Steel/Fighting x1
For a straight line, if A = (x1 , y1) and B = (x2 , y2) are any two points on the line, then the slope is (y2 - y1)/(x2 - x1) provided x2 is not the same as x1. More generally, if the equation is y = f(x) then the rate of change in y is dy/dx or f'(x), the derivative of the function f(x).