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Q: What is the integral of the derivative with respect to x of a function of x with respect to x?

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We say function F is an anti derivative, or indefinite integral of f if F' = f. Also, if f has an anti-derivative and is integrable on interval [a, b], then the definite integral of f from a to b is equal to F(b) - F(a) Thirdly, Let F(x) be the definite integral of integrable function f from a to x for all x in [a, b] of f, then F is an anti-derivative of f on [a,b] The definition of indefinite integral as anti-derivative, and the relation of definite integral with anti-derivative, we can conclude that integration and differentiation can be considered as two opposite operations.

âˆ« [f'(x)g(x) - g'(x)f(x)]/g(x)2 dx = f(x)/g(x) + C C is the constant of integration.

If x is a function of time, t, then the second derivative of x, with respect to t, is the acceleration in the x direction.

The derivative of cosine of x is simply the negative sine of x. In mathematical terms f'(x) = d/dx[cos(x)] = -sin(x)

If it is with respect to t: 1 If it is with respect to some other variable (x for example): (dt)/(dx), which is literally read "the derivative of t with respect to x"

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d/dx âˆ« f(x) dx = f(x)

âˆ« f'(x)/f(x) dx = ln(f(x)) + C C is the constant of integration.

âˆ« f'(x)g(x) dx = f(x)g(x) - âˆ« f(x)g/(x) dx This is known as integration by parts.

To integrate a function you find what the function you have is the derivative of. for example the derivative of x^2 is 2x. so the integral of 2x is x^2.

âˆ« f(x)nf'(x) dx = f(x)n + 1/(n + 1) + C n â‰ -1 C is the constant of integration.

What do you mean? As this is a calculus question, I presume that you are asking for a derivative or integral The derivative of any function of the form ƒ(x) = a * x ^ n is ƒ'(x) = a * n * x ^ (n-1) The integral of any function of the form ∫ a*x ^ n is a / (n+1) * x ^ (n+1) + C Your function that you gave is 1 / x^(2) which is equal to: x^(-2) Thus the derivative is: -2 * x^(-3) And the integral is: -x^(-1) + C

We say function F is an anti derivative, or indefinite integral of f if F' = f. Also, if f has an anti-derivative and is integrable on interval [a, b], then the definite integral of f from a to b is equal to F(b) - F(a) Thirdly, Let F(x) be the definite integral of integrable function f from a to x for all x in [a, b] of f, then F is an anti-derivative of f on [a,b] The definition of indefinite integral as anti-derivative, and the relation of definite integral with anti-derivative, we can conclude that integration and differentiation can be considered as two opposite operations.

âˆ« f'(x)/âˆš(af(x) + b) dx = 2âˆš(af(x) + b)/a + C C is the constant of integration.

The derivative with respect to 'x' is 4y3 . The derivative with respect to 'y' is 12xy2 .

âˆ« f'(x)/âˆš[f(x)2 + a] dx = ln[f(x) + âˆš(f(x)2 + a)] + C C is the constant of integration.

You can differentiate a function when it only contains one changing variable, like f(x) = x2. It's derivative is f'(x) = 2x. If a function contains more than one variable, like f(x,y) = x2 + y2, you can't just "find the derivative" generically because that doesn't specify what variable to take the derivative with respect to. Instead, you might "take the derivative with respect to x (treating y as a constant)" and get fx(x,y) = 2x or "take the derivative with respect to y (treating x as a constant)" and get fy(x,y) = 2y. This is a partial derivative--when you take the derivative of a function with many variable with respect to one of the variables while treating the rest as constants.

For positive x, this expression is equal to 1. The integral (anti-derivative) is therefore x + C (where C is the arbitrary integration constant). For negative x, this expression is equal to -1, and the integral is -x + C. Wolfram Alpha gives the integral as x times sgn(x), where sgn(x) is the "sign" function.

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