d/dx (X - 1)x
= (X - 1)x ln(X - 1) * x
= X(X- 1)x ln(X - 1)
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To find the derivatve of the square root of cos x: Use the chain rule; this means multiply the inner derivative by the outer derivative. You can write the question f(x) = (cos x)1/2 This general break-down explains how to find d/dx f(x) note: d/dx basically symbolizes "the derivative of" In general terms: f(x) = x1/2 g(x) = cos x f(g(x)) = (cos x)1/2 outer derivative: d/dx f(z) = (1/2)*x-1/2 = 1/(4cos x)1/2 inner derivative: d/dx g(x) = -sin(x) final answer: d/dx f(g(x)) = -sin(x)/(4*cos x)1/2 note: d/dx means "the derivative of"; so d/dx x = 1 Further explained: Set up the equation to a more general form: (cos x)1/2 To make the inner derivative, look at cos(x) To make the outer derivative, look at x1/2 note: x ~ cos x; so we treat (cos x) simply as x, to create the outer derivative You probably know the necessary derivates: 1. derivative of cos x = -sin x 2. derivative of a1/2 = (1/2)*a-1/2 = 1/(4a)1/2 Multiplying the two we get the answer: -sin(x)/(4cos x)1/2
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).
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
Any number raised to the power 1 is that same number, x1 = x. For example, 51 = 5.
The derivative with respect to 'x' of sin(pi x) ispi cos(pi 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)
x*x1/2= x3/2 Derivative = 3/2 * x1/2
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
Write square root of x as x1/2. Then use the formula for the derivative of a power.
The derivative is 2x based on the power rule. Multiply the power by the coefficient of x then drop the power by one.
-1
x1 = x
The first derivative of e to the x power is e to the power of x.
Suppose you have a differentiable function of x, f(x) and you are seeking the root of f(x): that is, a solution to f(x) = 0.Suppose x1 is the first approximation to the root, and suppose the exact root is at x = x1+h : that is f(x1+h) = 0.Let f'(x) be the derivative of f(x) at x, then, by definition,f'(x1) = limit, as h tends to 0, of {f(x1+h) - f(x1)}/hthen, since f(x1+h) = 0, f'(x1) = -f(x1)/h [approx] or h = -f'(x1)/f(x1) [approx]and so a better estimate of the root is x2 = x1 + h = x1 - f'(x1)/f(x1).
The derivative of ( x1/2 ) with respect to 'x' is [ 1/2 x-1/2 ], or 1/[2sqrt(x)] .
It is x1 which is x.
The derivative of ex is ex